F  ROM  -TH  E  -  L!  BRARY-  OF 
•W1LL1AM-A  HILLEBRAND 


• 

•••',;•;•.•.•. 


o 


NOTES  AND  EXAMPLES 


—  IN 


MECHANICS; 


WITH    AN    APPENDIX   ON   THE 


GRAPHICAL  STATICS  OF  MECHANISM ; 


—  BY 


IRVING  P.  CHURCH,  C.  E., 

Professor  of  Applied  Mechanics  and  Hydraulics  ,  College  of  Civil  Engineering, 
Cornell  University. 


SECOND  EDITION,  REVISED  AND  ENLARGED. 


NEW  YORK : 

JOHN  WILEY  &  SONS. 

LONDON  :  CHAPMAN  &  HALL,  LIMITED. 

1903. 


*  3 


COPYRIGHT,  1891, 

•• BY 

P.  CHURCH. 


ERRATA. 

P.  63.    In  Fig.  76  for  "  12.2'"  read  "  12.5'." 
P.  69.    I/ast  line  but  one,  for  "  4937  Ibs."  read  "  4973  Ibs." 
P.  70.    Second  line,  for  "  7063  Ibs."  read  "  7027  Ibs." 
Plate  V  of  Appendix  ;  in  Fig.  17  [A]  for  "  S  "  read  "  82 ' 


PEEFACE. 


THE  following  pages  form  a  companion  volume  to  the  writer's  Mechanics  of 
Engineering,  and  contain  various  notes  and  many  practical  examples,  both 
algebraic  and  numerical,  serving  to  illustrate  more  fully  the  application  of 
fundamental  principles  in  Mechanics  of  Solids  ;  together  with  a  few  paragraphs 
relating  to  the  Mechanics  of  Materials,  and  an  Appendix  on  the  "Graphical 
Statics  of  Mechanism  " 

Advantage  has  been  taken  of  the  use  of  preliminary  impressions  in  the 
classroom  to  make  corrections  in  the  electrotype  plates;  and  it  is  therefore 
thought  that  the  present  complete  edition  is  comparatively  free  from  typo- 
graphical errors. 

In  the  Appendix  are  presented  many  of  the  problems  of  Prof.  Herrmann's 
' '  Zur  Qraphischen  Statik  der  Maschinengetriebe  "  in  what  seems  to  the  writer 
a  clearer  form  than  in  the  original  (for  reasons  stated  on  the  first  page  of 
Appendix).  In  this  part  of  the  work,  the  text  and  diagrams  not  being  adjacent, 
alternate  pages  have  been  left  blank  in  such  a  way  that  any  diagram  and  its 
appropriate  text  can  be  kept  in  view  simultaneously. 

Besides  his  indebtedness  to  Prof.  Herrmann's  work,  the  writer  would  grate- 
fully acknowledge  the  kindness  of  the  Messrs.  Wiley  in  securing  a  higher 
order  of  excellence  in  the  execution  of  the  diagrams  than  had  at  first  been 
contemplated. 

In  references  to  the  writer's  Mechanics  of  Engineering  the  abbreviation 
M.  of  E.  is  used. 

CORNELL  UNIVERSITY, 
ITHACA,  N.  Y.,  March,  1892. 


PEEFACE   TO   SECOND   EDITION. 


FOB,  this  second  edition  the  plates  of  the  first  have  been  carefully  revised 
and  corrected  (except  as  indicated  in  the  errata  on  the  opposite  page),  and  an 
entirely  new  chapter  added  (Chap.  VIII,  pp.  119-133),  containing  various  notes 
and  explanations,  as  also  many  examples  for  practice. 

ITHACA,  January,  1897. 


995884 


CONTENTS. 


CHAP.  I.    DEFINITIONS.    PRINCIPLES.    CENTER  or  GRAVITY. 

PAGES 

§§  1-15.   Illustrations  of  Forces.   Mass,  Weight,  Equilibrium.   Fundamental 

Theorem  of  the  Integral  Calculus.    Centers  of  Gravity.   Simpson's  Eule.      1-J4 

CHAP.  II.    PRINCIPLES  AND  PROBLEMS  INVOLVING  NON- 
CONCURRENT  FORCES  IN  A  PLANE. 

§§  16-40a.  Conditions  of  Equilibrium.  Classification  of  Rigid  Bodies.  Two- 
force  Pieces ;  Three-force  Pieces,  etc.  Levers ;  Bell-crank ;  Cranes, 
Simple  and  Compound.  Eedundant  Support.  Two  Links ;  Rod  and 
Tumbler ;  Door ;  Wedge  and  Block.  Roof  Truss,  and  Cantilever  Frame.  15-42 

CHAP.  III.    MOTION  OF  A  MATERIAL  POINT. 

§§41-51.    Velocity.   Acceleration.    Momentum.    Cord  and  Weights.    Lifting     . 
a  Weight.    Harmonic  Motion.    Ballistic  Pendulum.    Balls  and  Spring. 
Cannon  as  Pendulum.    Simple  Circular  Pendulum 43-53 

CHAP.  IV.    NUMERICAL  EXAMPLES  IN  STATICS  OF  RIGID  BODIES 
AND  DYNAMICS  OF  A  MATERIAL  POINT. 

§§  52-72a.  Center  of  Gravity.  Toggle  Joint.  Crane.  Door.  Roof  Truss. 
Train  Resistance.  Motion  on  Inclined  Plane.  Block  Sliding  on  Circular 
Guide,  etc.  Harmonic  Motion  of  Piston.  Conical  Pendulum.  Weighted 
Governor.  Motion  in  Curve.  Motion  of  Weighted  Piston  with  Steam 
used  Expansively.  Ball  Falling  on  Spring 54-76 

CHAP.  V.    MOMENT  OF  INERTIA  OF  PLANE  FIGURES. 

§§  72b-76.  Section  of  I-beam  ;  of  Box-beam.  Irregular  Figures,  by  Simp- 
son's Rule.  Graphical  Method 77-82 

-.'   t  '•-•  CHAP.  VI.    DYNAMICS  OF  A  RIGID  BODY. 

§§  77-97.  Rotary  Motion.  Pendulum.  Speed  of  Fly-wheel.  Centrifugal 
Action  on  Bearings.  "  Centrifugal  Couple."  Piles.  Kinetic  Energy  of 
Rotary  Motion.  Work  and  Energy.  Numerical  Examples.  Action  of 
Forces  in  Locomotive.  The  Appold  and  Carpentier  Dynamometers. 
Boat  Rowing.  Solutions  of  Numerical  Examples.  Work  of  Rolling 
Resistance.  Strap  Friction  ;  Examples , 83-106 

CHAP.  VII.    MECHANICS  OF  MATERIALS  AND  GRAPHICAL  STATICS. 

§§  98-107.  Stresses  in  a  Rod  in  Tension.  Rivet-spacing  in  a  Built  Beam. 
I-beams;  without  using  Moment  of  Inertia.  "Incipient  Flexure  "in 
a  Column.  Tests  of  Wooden  Posts.  The  Pencoyd  Experiments  in 
Columns.  Graphical  Constructions 107-118 

CHAP.  VIII.     MISCELLANEOUS  NOTES. 

§§  108-116.  Center  of  Gravity.  The  Time- velocity  Curve.  Reduction  of 
Moment  of  Inertia  of  Plane  Figure.  Miscellaneous  Examples.  The 
"  Imaginary  System  "  in  Motion  of  Rigid  Body.  Angular  Motion 119-133 

APPENDIX  ON  THE  GRAPHICAL  STATICS  OF  MECHANISM.  1-34 

(See  p.  28  of  the  Appendix  for  its  Table  of  Contents.) 


NOTES  AND   EXAMPLES  IN  MECHANICS. 


CHAPTEE  I.   \  t  /;,  • 
DEFINITIONS.     PRINCIPLES.     CENTRE  OF  GRAVITY. 

1.  Applied  Mechanics  is  perhaps  a  more  common  term  for  the 
same  thing  than  u  Mechanics  of  Engineering."     "  Pure  Mechan- 
ics" is  another  name  for  Analytical  Mechanics,  which  deals  with 
the  subject  entirely  from  a  mathematical  point  of  view. 

2.  Abstract  Numbers. — In  experimental  investigations  in  which 
formulae  are  to  be  deduced,  it  is  best  to   throw   experimental 
coefficients  into  the  form  of  abstract  numbers,  if  possible,  for 
these  are  immediately  comparable  with  those  of  another  experi- 
menter in  the  same  field,  if  the  latter  follows  the  same  plan., 
whether  he  uses  the  same  units  for  space,  force,  and  time,  or  not. 

Thus :  if  the  coefficient  of  friction  be  defined  as  the  ratio  of 
the  friction  [force]  to  the  normal  pressure  [force]  producing  it, 
we  obtain  the  same  number  for  it  in  a  definite  experiment, 
whether  we  express  our  forces  in  pounds  or  in  kilograms. 

3.  Forces. — One  of  the  most  important  things  to  be  acquired 
in  dealing  with  the  practical  problems  of  this  study  is  a  proper 
conception  of  forces.     We  do  not  use  the  word  force  in   any 
abstract  general  sense,  nor  in  any  popular  sense,  such  as  is  in- 
stanced in  the  Note  of  §  15c,  M.  of  E.     It  should  always  mean 
the  pull,  pressure,  rub,  attraction  (or  repulsion),  of  one  body  upon 
another,  and  always  implies  the  existence  of  a  simultaneous,  equal, 
and  opposite  force  exerted  by  that  other  body  on  the  first  body, 
i.e.,  the  reaction  /  but  this  reaction  will  not  come  up  for  consid- 
eration in   any  problem  unless  this  "first"  body  is  under  treat- 


2  NOTES   AND   EXAMPLES   IN   MECHANICS. 

ment  as  regards  the  forces  acting  on  it.  In  most  problems  in 
Mechanics  we  have  one  or  more  definite  rigid  bodies  under  con- 
sideration, one  at  a  time,  in  whose  treatment  we  must  form  clear 
conceptions  of  the  forces  acting  on  it ;  and  these  always  emanate 
from  other  bodies. 

Hence  in  no  case  should  we  call  anything  a  force  unless  we, 
can  conceive  of  it  as  capable  of  measurement  by  a  spring-balance, 
•  &n&.are  able  do  say  from  what  other  ~body  it  comes. 

\  *>tFor  vx&mpl&,*&'\)ody  said  to  weigh  30  Ibs.  lies  at  rest  on  a 
;6&i£)<>th;i£Vel'  table,  which  is  the  only  body  with  which  it  is  in 
'c'orit'acft."  "'Wlten*  6ortsrdered  by  itself  this  body  is  acted  on  by  only 
two  other  bodies  in  a  manner  which  justifies  the  use  of  the  word 
force  ;  viz.,  the  action  of  the  earth  upon  it  is  a  vertical  downward 
attraction  (force)  of  30  Ibs. ;  while  the  action  of  the  table  upon  it 
is  an  upward  pressure  (force)  of  30  Ibs.     (We  here  ignore  the 
atmosphere  whose  pressures  on  the  body  are  balanced   in  every 
direction.)     But  suppose  the  same  body  and  the  table  with  which 
it  is  in  contact  to  be  allowed  to  fall,  from  rest,  in  a  vacuum.    The 
two  bodies,  during  the  fall,  remain  apparently  in  as  close  contact 
as  before ;  but  now  the  upper  body  is  under  the  action  of  only 
one  force,  viz.,  the  downward  attraction  of  the  earth,  30  Ibs. ;  and 
there  is  no  pressure  of  the  upper  body  against  the  table,  and  con- 
sequently no  pressure  of  the  table  against  the  upper  body. 

As  another  instance,  an  iron  rod  rests  horizontally  on  two 
level-faced  supports,  at  its  extremities,  and  bears  a  load  of  60  Ibs. 
in  the  middle.  "When  this  rod  is  considered  "free"  i.e.,  when  those 
other  bodies  which  act  on  it  in  a  u/bm?-able"  way  are  supposed 
removed  (their  places  being  for  present  purposes  taken  by  the 
respective  forces  with  which  they  act  on  the  first  body),  we  find 
it  to  be  under  the  action  of  four  forces,  viz. :  a  pressure  on  its 
middle,  vertical  and  downward,  of  60  Ibs.  from  its  load ;  the 
downward  attraction  of  the  earth  on  it,  i.e.,  its  own  weight,  say 
10  Ibs.  (which  is  really  distributed  among  all  of  its  particles,  but 
which,  so  far  as  the  equilibrium,  or  state  of  rest  or  motion,  of  the 
body  is  concerned,  is  the  same  as  if  applied  at  the  centre  of 
gravity,  viz.,  the  middle  of  the  rod) ;  and  the  two  upward  press- 
ures of  the  two  supports  against  the  ends  of  the  rod,  these  being 


FORCES. 


35  Ibs.  each.  If  the  nature  of  the  investigation  requires  it,  we 
may  go  on  and  consider  one  of  the  supports  by  itself,  or  "  free" ; 
in  which  case,  whatever  the  actions  of  other  bodies  on  it  may  be, 
that  of  the  rod  will  be  a  downward  force  of  35  Ibs.,  the  equal 
and  opposite  of  the  35  Ibs.  upward  pressure  of  the  support  against 
the  rod.  These  pressures  of  the  two  supports  against  the  rod  are 
usually  called  the  '•  Reactions  of  the  Supports." 

As  another  instance :  a  ball  of  10  Ibs.  weight  hangs  at  rest  by 
a  cord  attached  to  a  support  above.  The  cord  is  of  course  ver- 
tical. This  ball  is  under  the  action  of  two  forces,  viz.,  a  down- 
ward attraction  of  10  Ibs.  emanating  from  the  earth,  and  an 
upward  pull  of  10  Ibs.  emanating  from  the  cord. 

A  portion  of  the  above  cord,  taken  in  the  part  under  ten- 
sion, is  under  the  action  of  two  forces,  thus :  the  part  just 
above  it  exerts  an  upward  pull  of  10  Ibs.  upon  it,  and  that  below 
it  exerts  a  downward  pull  of  10  Ibs.  upon  it.  (We  here  neglect 
the  weight  of  the  portion  of  cord  considered  as  presumably  very 
small.)  In  such  a  case  the  tension  of  the  cord  is  said  to  be  10 
Ibs.  (not  20  Ibs.). 

Further  illustration.  Fig.  1  shows  a  prismatic  rod  CB  lean- 
ing against  the  smooth  vertical  side  of  a  block.  Both  rest  on  a 
rough  horizontal  plane.  The  rod  is  under 
three  forces,  viz. :  its  weight  G  acting 
vertically  downwards  through  its  middle; 
the  pressure  of  the  wall  against  it,  P 
(which,  since  the  wall-surface  is  perfectly 
smooth,  must  be  horizontal  and  points 
toward  the  right) ;  and  a  third  force,  Q, 
the  pressure  of  the  floor  against  the  rod. 
Since  the  rod  and  block  are  at  rest,  P  and 
G  intersecting  at  J.,  it  must  be  that  the  floor  is  sufficiently  rough 
to  enable  the  pressure  Q  to  deviate  from  the  vertical  (that  is,  from 
the  normal  to  plane  of  floor)  by  as  much  as  the  angle  AB  F,  at 
least;  for,  as  will  be  proved  later,  if  three  forces  act  on  a  body 
and  it  remains  at  rest,  the  three  lines  of  action  must  intersect  in 
a  common  point. 

We  next  consider  the  block,  or  wall,  by  itself,  and  find  it  to 


FIG. l. 


4  NOTES   AND    EXAMPLES   IN   MECHANICS. 

be  under  the  action  of  P' ,  the  equal  and  opposite  of  jP,  and 
therefore  pointing  horizontally  toward  the  left ;  of  6r',  its  weight; 
and  a  pressure,  R,  from  the  floor,  whose  line  of  action  is  deter- 
mined by  the  fact  that  it  must  be  the  same  line  of  action  as  that  of 
the  (ideal)  resultant  of  the  forces  G-'  and  P\  since  if  three  forces 
balance,  i.e.,  are  in  equilibrium,  any  one  of  them  must  be  the- 
equal  and  opposite  of  the  resultant  of  the  other  two  and  have  the 
same  action-line.  Forming  a  parallelogram,  therefore,  on  the 
forces  P'  and  6?',  first  conceiving  each  of  the  two  forces  to  be 
transferred  in  its  line  of  action  to  their  point  of  intersection,  A', 
the  diagonal  of  this  parallelogram  represents  the  equal  and  oppo- 
site of  ^,  and  has  the  same  line  of  action. 

If  this  diagonal  intersects  the  floor  on  the  left  of  the  lower 
left-hand  corner  of  the  block,  the  supposed  stability  is  impossible, 
unless  the  block  is  cemented  to  the  floor. 

4.  Mass  and  Weight. — The  question  of  mass  will  be  further 
discussed  in  a  subsequent  chapter,  p.  53,  M.  of  E.    By  weight  we 
are  always  to  understand  the  force  of  the  attraction  which  the 
earth  exerts  upon  the  body,  and  not  the  amount  of  matter  (mass) 
in  it.    This  weight  will  therefore  be  different  in  different  latitudes 
and  at  different  distances  from  the  centre  of  the  earth,  and  re- 
quires a  spring -balance  for  its  determination.    Physicists  also  use 
the  word  weight  in  this  sense  (force). 

5.  The  Heaviness  of  a  body,  in  the  sense  used  in  this  study,  is 
something  quite  different  from  its  total  weight.     For  instance,  if 
the  substance  of  the  body  is  not  uniform  in  composition  and  den- 
sity, we  cannot  speak  of  the  heaviness  of  the  body  as  a  whole, 
since  its  various  portions  have  not  a  common  heaviness  ;  however, 
we  may  speak  of  its  average  heaviness,  which  might  be  of  some 
use  in  certain  problems,  and  would  be  the  quotient  of  its  total 
weight  divided  by  its  volume. 

Since  heaviness  is  not  an  abstract  number,  it  would  not  be 
sufficient  to  say  that  a  certain  substance  has  a  heaviness  of  40, 
for  instance,  nor  even  40  Ibs. ;  the  full  statement  must  be  40  Ibs. 
per  cubic  foot ;  which  is  equivalent  to  the  statement  that  the 
heaviness  is  0.540  ton  per  cubic  yard. 

6.  Rigid  Body. — As  an  illustration  of  the  definition  in  §  10, 


EQUILIBRIUM — TRANSMISSIBILITY    OF  FORCE.  5 

M.  of  E.,  it  may  be  said  that  if  a  horizontal  bar,  supported  at  its 
extremities,  is  so  moderately  loaded  that  the  deflection  or  sinking 
of  the  central  point  is  only  about  one  3-hundredth  part,  for  in- 
stance, of  the  span  or  distance  between  the  supports,  it  is  suffi- 
ciently accurate,  for  most  purposes,  to  consider  that  there  has  been 
no  change  in  the  length  of  the  horizontal  projection  of  any  dis- 
tance measured  along  the  bar.  (Where  such  a  consideration  is 
inadmissible,  attention  will  be  called  to  it.) 

7.  Equilibrium. — Besides  speaking  of  a  system  of  forces  being 
in  equilibrium,  the  phraseology  is  also  sometimes  used  that  the 
rigid  body  is  in  equilibrium  under  the  forces  acting  on  it  (§  40a). 

The  reservation  made  in  §  11,  M.  of  E.,  as  to  state  of  motion 
refers  to  the  fact  that  any  alteration  in  the  distribution  of  forces 
acting  on  a  rigid  body  will  usually  cause  a  difference  in  the  inter- 
nal strains  and  stresses  produced,  though  the  state  of  motion  may 
or  may  not  be  affected,  according  as  any  second 

system  of  forces  applied  to  the  body,  on  removal  of  the  first,  has 
a  different  resultant  from  that  of  the  first,  or  the  same  resultant. 

8.  Division  of  the  Subject. — As  to  the  division  given  in  §  12, 
M.  of  E.,  Sir  William  Thomson,  the  noted  English  physicist,  has 
adopted  a  different  nomenclature,  which  is  getting  into  wider  and 
wider  use.     He  makes  the  term  Dynamics  include  both  statics 
and  dynamics  (i.e.,  what  is  here  and  by  Rankine  and  Continental 
writers  called  Dynamics),  and  replaces  their  word  Dynamics  by 
Kinetics. 

9.  Transmissibility  of  Force.    Resultant. — The  principle  of  the 
transmissibility  of  force  refers  only  to  the  state  of  motion  of  the 
rigid  body.    For  instance  (see  Fig.  2),  as  far  as  the  rest  or  motion 
of  the  sickle-shaped  body  is  concerned,  it  is 

immaterial  whether  the  force  P  balance  P' 
(being  equal  and  opposite  to  it  and  in  the 
same  line)  by  being  applied  at  0,  or  by  being 
applied  at  A  ;  but  in  the  former  case  the  part 
ABO  would  be  under  a  bending  strain,  and  in 
the  latter  would  be  under  no  strain. 

It  must  be  remembered  that  the  resultant  of  a  given  system 
of  forces  is  always  a  purely  imaginary  force ;  that  is,  all  we  mean 


6  NOTES   AND    EXAMPLES   IN   MECHANICS. 

is  this :  that  if  the  given  forces  were  removed  and  their  resultant 
acted  in  their  stead,  in  proper  position  as  well  as  magnitude,  the 
state  of  motion  of  the  rigid  body  would  not  be  any  different  from 
what  it  would  be  without;  the  replacement. 

10.  Parallelogram  of  Forces;  or  Triangle  of  Forces. — By  some 
it  is  contended  that  this  geometrical  construction,  or  relation,  the 
Parallelogram  of  Forces,  should  give    place  to  the  triangle  of 
forces ;  i.e.,  that  the  resultant  (a  line  laid  off  to  scale  representing  it) 
is  equal  to  the  third  side  of  a  triangle  whose  other  two  sides  are  the 
two  given  forces ;  but  a  construction  of  that  nature  does  not  show 
the  resultant  as  acting  through  the  same  point  as  the  two  com- 
ponents, which  should  be  the  case  if  the  construction  is  to  give 
the  position,  as  well  as  the  magnitude  of  the  resultant. 

It  is  true  that  the  systematic  application  of  the  triangle  of 
forces  gives  rise  to  the  methods  of  Graphical  Statics,  as  will  be 
seen,  but  in  that  case  the  magnitudes  of  all  the  forces  (or  rather 
lines  of  definite  length  representing  them  and  parallel  to  them) 
are  drawn  on  a  separate  part  of  the  paper  from  that  containing 
the  figure  of  the  body  acted  on. 

11.  General  Remarks  on   Forces. — It  is   not  such   a   simple 
matter  as  it  might  at  first  appear,  to  bring  to  bear  upon  a  given 
body  a  force  of  prescribed  magnitude  and  direction  at  a  specified 
point.    For  example,  if  we  have  the  idea  that  a  given  tension  can 
be  produced  in  a  vertical  cord  by  hanging  upon  it  a  body  whose 
weight  is  the  given  amount,  we  must  remember  that  the  tension 
in  the  cord  will  be  equal  to  the  body's  weight  only  in  case  the 
body  is  at  rest  or  moving  in  a  right  line  with  unchanging  velocity 
(i.e.,  describing  equal  spaces  in  equal  times). 

While  we  can  always  be  sure  that  the  weight  of  the  body  itself 
(or  action  of  the  earth  on  it)  is  <i  force  of  constant  value  acting 
on  it  at  all  times  and  in  a  vertical  downward  direction,  the  values 
of  the  pressures  or  pulls  exerted  upon  it  by  neighboring  bodies 
depend  on  the  state  of  motion  of  the  bodies  concerned,  as  well  as 
on  their  weights.  For  instance,  suppose  that  we  wish  to  observe 
the  effect,  upon  a  block  of  metal  placed  on  a  smooth  level  table 
and  weighing  32.2  Ibs.,  of  bringing  to  bear  upon  it  a  horizontal 
force  of  10  Ibs.  If  (see  Fig.  3)  we  attach  to  it  a  light  flexible 


THEOREM    OF   THE   INTEGRAL    CALCULUS. 


FIG.  3. 


(but  inextensible)  cord,  led  off  horizontally  to  the  right  and  then 

passing  over  a  frictionless  and  very  light 

pulley,   M,   and   fasten  a  10-lb.    weight, 

B,  to  the  end  of  the  right-hand  vertical 

portion,  we  find,  by  interposing  a  light 

spring-balance  between  the  two  parts  of 

the    cord    at    A,  that    when    motion   is 

allowed  to  begin  (the  cord  being  pulled 

straight  before  the  right-hand  weight  is  allowed  to  sink  freely), 

the  tension  at  A  is  not  10  Ibs.,  but  only  7.63  Ibs. 

If  now  we  gradually  increase  the  weight  at  B  in  successive 
experiments  until  the  spring-balance  in  the  cord  at  A  shows  a 
reading  of  just  10  Ibs.  (the  desired  force),  we  find  that  the  weight 
at  B  has  reached  a  value  of  14.5  Ibs.  (The  reasons  for  this  will 
appear  in  Ex.  2  of  §  57,  M.  of  E.) 

The  effect  of  the  10-lb.  force  on  the  block  is  then  noted  to  be 
as  follows :  The  block  begins  to  move  toward  the  right  (having 
been  initially  at  rest)  with  an  accelerated  motion.  In  the  first 
second  it  passes  over  5  ft. ;  in  the  second  second,  15  ft. ;  in  the 
third,  25  ft. ;  and  so  on,  as  the  odd  numbers,  1,  3,  5,  etc.  In  other 
words,  the  total  distance  from  the  start  is  equal  to  5  ft.  X  the 
square  of  the  total  time  in  seconds. 

12.  Review  of  the  Fundamental  Theorem  of  the  Integral  Cal- 
culus.— This  must  be  thoroughly  reviewed  and  understood  before 
using  the  integral  calculus  in  the  theory  of  the  centre  of  gravity, 

or  elsewhere  in  Mechanics.  The 
problems  to  which  we  apply  the 
integral  calculus  will  almost  always 
be  of  a  nature  calling  for  the  sum- 
ming up  of  a  vast  number  of  very 
small  quantities  all  of  which  are 
alike  in  form  and  usually  consist 
of  the  product  of  two  or  more  fac- 
tors, one  of  which  [the  differential} 
is  very  small,  from  which  it  comes 
that  the  product  itself  is  small.  (This  is  the  most  available  form 
in  which  to  conceive  of  the  operation,  but,  as  will  be  seen,  it  is 


8  NOTES   AND   EXAMPLES   IN   MECHANICS. 

after  all  a  fictitious  description  of  what  is  done,  being  put 
into  that  shape  to  avoid  the  continued  repetition  of  a  very 
lengthy  phraseology.)  This  theorem  and  operation  are  most 
readily  appreciated  by  the  aid  of  graphic  representation,  as 
follows  : 

Let  ON  be  any  portion  of  any  algebraic  curve,  its  equation . 
being 

y  —funG.  of  x ;     or,    y  —  0(a?) (1) 

At  any  point  m  of  the  curve  there  is  a  special  value  of  x  and 
of  y,  and  also  of  the  angle  <*,  which  the  tangent-line  there  drawn 
makes  with  the  axis  X. 

From  the  Differential  Calculus  we  know  that  tan  a  is  the  "  first 
differential  coefficient "  of  y  with  respect  to  x  (or  "  first  deriva- 
tive," or  simply  "  derivative"),  and  may  be  written  0'(a?) ;  whence 
with  symbols, 

ii ?/ 

tan  a  = ;/-;    or    tan  a  =  0'(a?).     ...     .     (2) 

Now  divide  the  distance  A  .  .  .  B  into  a  large  number  of 
parts,  not  necessarily  equal,  the  length  of  any  one  being  called 
Ax,  and  raise  an  ordinate  at  each  point  of  subdivision.  Where 
each  such  ordinate  cuts  the  curve,  as  at  m,  draw  a  tangent  line  to 
the  curve,  and  a  horizontal  line  as  m  .  .  .  o.  These  two  lines  in- 
tercept a  small  length  a  .  . .  b  on  the  next  ordinate  on  the  right. 
All  such  intercepts  are  shown  drawn  in  heavy  lines,  ab  is  typical 
of  any  one  of  these  intercepts.  From  the  right  triangle  concerned 
we  now  have 

ab  =  mb  .  tan  a\  i.e.,  ab  =  (tan  a)  Ax  =  \$>'(x}\Ax.  .     (3) 

Project  all  the  lengths  like  ab  on  a  convenient  vertical  line, 
as  ED,  and  note  that  their  sum  is,  of  course,  not  quite  equal  to 
ED,  or  NC,  which  is  yn  —  y<>>  This  sum  we  may  express  as 
and  hence  state  that 

is  almost  equal  to  yn  — y0.       .     .     (4) 


THEOREM  OF  THE  INTEGRAL  CALCULUS.         9 

Or,  since   yn  —  y<>   maJ  De   written    \4>(x)]x  =  Xn  —  [4>(x]\x=X(^ 

=       0(#),  we  may  re-state  the  fact  thus : 

LO 

rxn 
^N\_<p'(x)\dx  almost  =         00)-    »     „     0     „     (5) 

Since  it  is  evidently  a  geometrical  possibility  to  increase  the 
number  of  ^a?'s  between  A  and  B  until  the  discrepancy  between. 

^N\(J)'(x)~\Ax  and       "00)  is  less  than  any  value  that  maybe  as- 

r*n 

signed,  we  may  say  that         00)  is  the  limit  which  ^  r0'(a?)_WaJ 

Lo-0 

approaches  as  the  subdivision  becomes  finer  and  finer.  This  limit 
may  also  be  expressed  by  the  notation  /  n  Vcf)f(xf\dx,  and  we 
accordingly  write 


(  limit  called 

which  limit  —  I    '"  0(a?).     .     (6) 


The  utility  of  all  this  is  that  if  in  any  problem  we  note  that 
the  sum  of  a  series  of  terms,  each  one  of  which  is  the  product  of 
a  small  portion,  Ax,  of  the  axis  of  JJTby  a  function  of  x  (the  same 
function  in  all)  (Ax  being  the  difference  between  the  value  of  x 
for  any  term  and  that  for  the  next  term)  is  an  approximation  to 
the  desired  result ;  and  if,  furthermore,  the  desired  result  is  the 
limit  approached  in  value  by  the  sum  of  the  series  as  the  sub- 
division becomes  finer  and  finer  along  the  axis  of  X,  then  the 
desired  result  will  be  obtained  as  follows :  Find  the  anti-deriva- 
tive of  the  above  function  of  a?,  and  substitute  in  it  first,  for  a?,  the 
value,  xn ,  which  x  assumes  at  the  upper  end  of  the  series ;  and 
secondly,  substitute  the  value  xa  of  x  at  the  lower  end  of  the 
series.  The  difference  of  the  expressions  so  obtained  is  the  desired 
result.  (JSToTE. — By  anti-derivative  we  are  to  understand  that 
function  of  a?,  the  differentiation  of  which  will  give  rise  to  the 
given  function.  Thus,  the  anti-derivative  of  0/(o?)  =  a?2  ie 
0(a?)  =  %x*  +  const. ;  and  in  applying  the  above  rule  we  may 


10  NOTES   AND   EXAMPLES   IN   MECHANICS. 

omit  the  constant,  knowing  that  it  would  cancel  out  in  the  sub- 
traction indicated.) 

13.  Example  of  the  Foregoing. — We  wish  to  find  by  calculus 
the  total  moment  of  the  homogeneous  paraboloid  of  revolution  in 
* — "*K         ^£8  **  about  the  axis  Y\  the  axis  of  the 
solid  being  horizontal  and  coinciding  with 
the  axis  X.     (By  "moment"  of  a  small 
weight  we  mean  the  product  of  the  weight 
by  the  length  of  the  perpendicular  let  fall 
upon  it  from  a  given  line  or  plane).     The 
FIG.  5.  solid  is  bounded  on  the  right  by  a  circular 

base,  KJN,  perpendicular  to  axis  X.  Conceiving  the  solid  to  be 
divided  into  a  great  number  of  circular  disks,  or  laminae,  perpen- 
dicular to  axis  X,  all  of  the  same  thickness,  Ax,  but  of  different 
radii,  u,  and  denoting  the  "heaviness"  of  the  substance  by  y,  we 
have  yrtv?Ax  as  the  weight  of  any  disk,  and  yiru*Ax  [x  +  ^Ax~\ 
as  its  moment  about  the  axis  Y.  (The  axis  l^is  perpendicular 
to  the  paper  through  O.)  The  equation  to  the  parabola  being 
u*  =  %px,  where  p  is  the  parameter,  this  moment  may  also 
be  written  %ynpxAx\x  4-  \^x~\,  and  hence  the  sum  of  all  such 
moments  for  all  the  disks,  or  the  total  moment,  may  be  expressed 
as 

.       M'  =  Zy7tp^\_(tf)\Ax  +  ynpAx^\(x)\Ax.  .     .     (7) 

This  sum  does  not  apply  to  the  paraboloid,  but  only  to  the 
stepped  solid  (or  aggregation  of  disks  with  square  edges),  unless  Ax 
is  finally  made  zero.  That  is,  the  desired  result  (moment  of  actual 
paraboloid)  will  be  obtained  as  the  limit  which  M '  approaches 
as  Ax  is  made  smaller  and  smaller.  When  Ax  is  made  zero, 

the  first  term  of  Mf  becomes  %y7tp   I   n  (x*}dx ;  (i.e.,  <j>'(x)  —  x*) ; 

while   as   to   the   second,    although    the    limit   of   2N(x)Ax  is 

/   n(x)dx,  and  hence  is  not  zero,  yet  the  outside  factor,  Ax,  is 

t/X0 

zero  and  hence  the  second  term  vanishes.  Therefore  the  total 
moment  of  the  paraboloid  is  M  =  2y?rp  f  Xn\x*~]dx',  and  in  this 


CENTRE   OF   GRAVITY.  11 

we  note  that  the  <t>\x)  of  the  general  form  of  eq.  (6)  is  a?a,  and 
that  the  x  anti-derivative  of  a?2  is  Ja?3  +  const. ;  and  hence,  finally, 

M  = 

This  quantity  divided  by  the  total  weight  (=  ynpx^)  of  the  solid 
will  give  the  distance  of  its  centre  of  gravity  from  the  vertex,  or 
origin,  0 ;  i.e.,  x  =  %xn. 

As  to  notation,  it  is  customary  to  anticipate  the  fact  that  the 

desired  result  justifies  the  use  of  the  notation   /   n  \(f)f(x)\dx,  and 

t/X0 

to  employ  at  once  dx  for  Ax  in  making  out  the  form  of  one  term 
of  the  series,  dx  is  then  called  an  infinitesimal,  which  simply 
means  that  Ax  is  finally  to  become  zero;  in  other  words,  that  the 
result  sought  is  the  limit  which  the  sum  of  the  series  approaches 
as  Ax  diminishes. 

14.  Position  of  Centre  of  Gravity  of  Various  Geometrical  Forms. 
(Homogeneous,  etc.;  the  plane  figures  representing  thin- plates 
of  uniform  thickness.) 

Obelisk.  Fig.  6  shows  a  homogeneous  obelisk,  or  solid 
bounded  by  six  plane  faces,  of  which  two  are  rectangular  (hori- 
zontal in  this  figure)  with  cor- 
responding edges  parallel  (and 
hence  these  rectangular  faces 
are  parallel,  and  may  be  con- 
sidered as  bases).  Required 
the  distance,  z,  of  the  centre 
of  gravity,  C' ,  of  the  obelisk, 
from  the  base  EEGF.  See 
Fig.  6  for  notation,  h  is 
the  perpendicular  distance  be- 
tween the  bases. 

By  passing  a  plane  through  the  edge  AD  \\  to  face  BCGF;  a 
plane  through  AB  \\  to  DCGH\  and  noting  their  intersections 
(dotted  lines)  with  the  faces  of  the  obelisk,  we  subdivide  it  into 
the  following  geometric  forms : 

A  parallelepiped  ABCD-JMGL,  of  volume  F,  =  blh  and 


12 


NOTES    AND   EXAMPLES    IN   MECHANICS. 


having  its  centre  of  gravity  at  a  distance  zl  =  ^h  above 

base  1-1F\ 
A  triangular  prism  AD-IJLII,  of  volume  =  Yt  =  i&(^  —  l)h 

and  for  whose  centre  of  gravity  z^  —  -JA ; 
Another  triangular  prism  AB-KFMJ,  of  volume  =   F3  = 

\l(bi  —  b)/i  and  for  whose  centre  of  gravity  zs  =  -J/< ;  and 

finally, 
K pyramid  A-EKJI,  whose  volume  is  F4  —  J-A(51  —  5)(^  —  I) 

and  whose  u  mean  &"  is  ^4  =  \h. 

Hence  by  eq.  (3)  of  p.  19,  M.  of  E.,  with  z^  etc.,  instead  of 
?x ,  etc.,  w^e  have,  after  reduction, 

77          I        O77l       77l        77  7 

—         o.L  -+-  6t)L  -4-  O.I  H-  Off,       fl 

11        I  II  II 


z  •=. 


vbl,  +  m  +  1,1  + 


Triangular  Plate.  Fig.  7.  Bisect  AB  in  M.  Join  6>J/. 
Bisect  0^  in  JV  and  join  JL^V.  The  intersection, 
O.  is  the  centre  of  gravity  of  the  triangular  plate  or 
plane  figure.  [The  centre  of  gravity  of  the  mere 
perimeter  of  the  triangle  (slender  wires,  homo- 
geneous and  of  same  sectional  area)  is  the  centre  of 
the  circle  inscribed  in  a  triangle  formed  by  joining 
the  middles  of  the  sides.] 

Parabolic  Plate.     Fig.    8.     AN  being   -j    to 
axis  ON.     ~ 


Upper  Half  of  preceding  Parabolic  Plate.    Fig.  9.    x— 
and  y  =  %AN. 


FIG.  8. 


FIG.  9. 


FIG.  10. 


—        $ 

/Semi-ellipse.    Fig.  10.    Semi-axes  are  a  and  I-  x  —  5—. 

O/T 

Sector  of  a  Sphere.  AEBO,  Fig.  11.  Let  h  =  the  altitude 
of  the  zone,  or  cap,  of  the  sector ;  i.e.,  let  h  —  r  —  OK\  then 
OC  :=  $r  —  -fA ;  C  being  the  centre  of  gravity. 


SIMPSON'S   EULE. 


13 


Segment  of  a  Sphere.     AK,  Fig.  12.     Let  the  altitude, 
of  the  segment  be  A,  and  C the  centre  of  gravity;  then 

(Zr  - 


Solid 
Spherical 
Segment. 


FIG.  12. 


FIG.  13. 


Join  the  vertex  0  with 
MN.     From  D  lay  off 


FIG.  14. 


Any  Pyramid  (or  Cone).     Fig.  13. 

Z>,  the  centre  of  gravity  of  the  base 

CD  =  fDO.      C  is   the   centre   of 

gravity  of  the  solid. 

Zone  on  Surface  of  Sphere.    Fig. 

]4.     (Thin   shell,  homogeneous  and 

of  uniform  thickness.)     The  centre 

of  gravity  lies  at  the  middle  of  the 

altitude,  A,  in  the  axis  of  symmetry.     The  small  circles  of  the 

sphere,  CD  and  AB.  lie  in  parallel  planes. 

15.  Simpson's  Rule   (Fig.  15).— If  ABCDEFG  is  a  smooth 

curve  and  ordinates  be 
drawn  from  its  extremities 
A  and  G  to  the  axis  X,  an 
approximation  to  the  value 
of  the  area  so  enclosed, 
A..D..G..N..O..A, 
between  the  curve  and  the 
axis  X,  is  obtained  by 
Simpsons  Rule,  now  to  be 
demonstrated.  Divide  the 

base  ON  into  an  even  number,  n,  of  equal  parts,  each  =  Ax  (so 

that  ON=n.4x),  and  draw  an   ordiriate  from  each   point  of 

division  to  the  curve,  the  lengths  of  these  ordinates  being  ul9 

u^ ,  etc. ;  see  figure. 


14  NOTES   AND   EXAMPLES   IN   MECHANICS. 

Consider  the  strips  of  area  so  formed  in  consecutive  pairs  ; 
for  example,  CDEE"  C"  is  the  second  pair  in  this  figure  (count- 
ing from  left  to  right).  Conceive  a  parabola,  with  its  axis  vertical, 
to  be  passed  through  the  points  C\  D,  and  E.  It  will  coincide 
with  the  real  curve  between  C  and  E  much  more  closely  than 
would  the  straight  chords  CD  and  DE\  and  the  segment  CDE^ 
considered  as  the  segment  of  this  parabola,  has  an  area  equal  to 
two  thirds  of  that  of  the  circumscribing  parallelogram  CO'EE. 
Hence,  since  the  area  of  this  pair  of  strips  =  trapezoid 
CEE"  C"  -{-parabolic  segment  CDE,  we  may  put 

tfltips  PCE>  \ 
which  reduces  to         ...  \Ax\u^  +  4ws  -f-  ^J- 

Treating  all  the  -^  pairs  of  strips  in  a  similar  manner,  we  have 

finally,  after  writing  Aw  —  (xn  —  o?0)  -=-  n, 

Whole  area  }  xn 
AG"  (appro*.}  \  = 
The  approximation  is  closer  the  more  numerous  the  strips  and 
the  more  accurate  the  measurement  of  the  ordinates  u^  ul9  u9,  etc. 
If  the  subdivision  on  the  axis  JT  were  "  infinitely  small,"  an 
exact  value  for  the  area  would  be  expressed  by  the  calculus  form 


/•a:  =  X 

I     ud 

*J  x  —  XQ 


udx.     Hence  for  any  integral  of  this  form,          udx,  if  we 


are  only  able  to  determine  the  particular  values  (u0  ,  ut  ,  etc.)  of 
the  variable  u  corresponding  respectively  to  the  abscissae  a?0, 
x0  -\-  Ax  ,  a?0  -j-  2^/aj,  etc.  (where  Ax  =  (xn  —  a?0)  -f-  n,  n  being  an 
even  number),  we  can  obtain  an  approximate  value  of  the  integral 
or  summation  by  writing 


As  to  the  meaning  of  n>  note  that  the  first  ordinate  on  the 
left  is  not  u^  but  u0 ;  also  that  while  there  are  n  strips,  the  num- 
ber of  points  of  division  is  n  -j-  1,  counting  the  extremities  O 
and^T. 


FORM  FOR  ANALYTICAL  CONDITIONS   OF   EQUILIBRIUM.    15 


CHAPTER  II. 

PRINCIPLES  AND  PROBLEMS  INVOLVING  NON-CONCURRENT  FORCES 

IN  A  PLANE. 

16,  Most  Convenient  Form  for  Analytical  Conditions  of  Equi- 
librium (Fig.  16). — Let  P, ,  P2 ,  P8 ,  etc.,  constitute  a  system  of 
non-concurrent  forces  in  a  plane  acting 
on  a  rigid  body  and  in  equilibrium.  Of 
the  actual  system,  P,  and  P2  are  the 
only  forces  shown  in  the  figure.  As- 
suming a  convenient  origin,  0,  introduce 
into  the  system  two  opposite  and  equal 
forces,  P/  and  P/',  both  acting  at  0  and 
equal  and  parallel  to  Pt .  Evidently  the 
presence  of  these  two  forces  does  not 
destroy  the  equilibrium  of  the  original 
system.  Similarly,  introduce  at  0  the  mutually  annulling  forces 
jy  and  P2"  bearing  the  same  relation  to  P2  (parallelism  and 
equality)  that  P/  and  P/'  do  to  P,  ;  and  so  on  for  each  of  the 
remaining  forces  of  the  system.  Drop  a  perpendicular  from  0 
on  each  of  the  forces  of  the  original  system,  the  lengths  of  these 
perpendiculars  being  a} ,  aa ,  a3 ,  etc.  (al  and  &2  are  shown  in  the 
figure).  We  now  note  that  for  each  force  P  of  the  original  sys- 
tem we  have  in  the  new  system  a  single  force  at  0,  equal  and 
parallel  to  P  and  similarly  directed,  and  also  a  couple,  of  moment 
Pa.  For  example,  the  force  P2  of  the  original  system  is  now 
replaced  by  the  force  P3"  parallel  and  equal  to  P,  and  similarly 
directed,  but  acting  at  the  point  0 ;  and  by  the  couple  formed  of 
the  two  forces  P2  and  P2',  the  arm  of  this  couple  being  a2.  It 
follows,  therefore,  that  the  new  system  consists  of  a  set  of  forces 
(P/',  P/',  P,",  etc.),  all  meeting  at  O  (and  hence  forming  a 
concurrent  system  in  a  plane  *),  and  a  set  of  couples,  of  moments 
Plal ,  Pa#a ,  etc.  Since  no  single  force  can  balance  a  couple 
(§  29,  M.  of  E.)  or  set  of  couples,  the  forces  of  the  concurrent 
system  at  0  must  be  in  equilibrium  among  themselves ;  i.e. , 
Y  being  any  two  directions  at  right  angles,  we  must  have 

*  And  therefore,  (unless  balanced,)  equivalent  to  a  single  force  or  resultant; 
see  M.  of  E.,  p.  8. 


16  NOTES   AND    EXAMPLES   IN   MECHANICS. 

and  2Y  separately  equal  to  zero  for  the  concurrent  system  at  0 ; 
and  the  set  of  couples  must  be  in  equilibrium  among  themselves, 
whence  it  follows  that  the  moment  of  their  resultant  couple  must 
equal  zero.  Since  the  couples  are  in  the  same  plane,  the  moment 
of  their  resultant  couple  is  the  algebraic  sum  (see  §  34,  M.  of  E.), 
i.e.,  Pfl,  +  P,a,  +  . . . ,  or  2(Pa),  =  0. 

For  the  equilibrium^  therefore,  of  a  system  of  non-concurrent 
forces  in  a  plane,  we  must  have  not  only  2X  and  2  Y  =  0,  but 
also  2(Pa)  =  0.  That  is,  in  practical  language  (the  body  being 
originally  at  rest),  the  forces  of  the  system  so  neutralize  each 
other  that  they  not  only  do  not  tend  to  move  the  body  sideways 
or  vertically,  but  also  do  not  produce  rotation. 

In  the  practical  application  of  these  conditions  of  equilibrium 
in  solving  problems  it  is  not  necessary  to  introduce  the  pairs  of 
equal  and  opposite  forces  at  0  (the  conception  of  which  is  needed 
only  for  purposes  of  proof),  since  the  sum  of  the  X-components 
(or  ^-components)  of  the  actual  forces  of  the  system  is  equal  to 
that  of  the  Jf-components  of  the  auxiliary  forces  introduced  at 
O\  while  to  form  the  moment-sum  of  the  auxiliary  couples,  we 
have  only  to  multiply  each  force  of  the  actual  system  by  the  per- 
pendicular distance  of  its  line  of  action  from  the  origin,  whose 
position  is  taken  at  convenience. 

The  student  should  now  read  the  latter  half  of  p.  33,  M.  of  E. 

17.  The  Rigid  Bodies  Dealt  with  at  Present. — Each  rigid  body 
now  to  be  considered  is  one  whose  dimensions  perpendicular  to 
the  paper  are  supposed  to  be  very  small,  and  therefore  may  be 
considered  to  lie  in  the  plane  of  the  paper.  An  actual  structure 
is  made  up  of  such  pieces,  or  members,  which  are  provided  with 
forked  joints,  or  duplicated  in  such  a  way  that  the  above  supposition 
(each  piece  lying  in  the  plane  of  the  paper)  is  practically  justified. 

All  surfaces  of  contact  between  any  two  contiguous  pieces  are 
supposed  perpendicular  to  the  paper,  and  friction  between  two 
such  parts  of  a  structure  is  disregarded  ;  i.e.,  the  pressure  between 
two  contiguous  pieces  (or  "  members")  is  in  the  plane  of  the  paper 
and  normal  to  the  surfaces  of  contact  /  for  it  is  a  matter  of  com- 
mon experience  that  pressure  can  be  exerted  at  the  smooth  sur- 
faces of  contact  of  two  bodies  only  in  a  direction  normal  to  those 
surfaces. 


CLASSIFICATION    OF   EIGID   BODIES.  17 

In  problems  where  the  weights  of  one  or  more  bodies  con- 
nected with  the  structure  are  considered,  the  plane  of  the  struc- 
ture will  be  vertical,  and  then  (considering  what  has  already  been 
postulated)  the  system  of  forces  acting  on  each  member,  or  piece 
of  the  structure,  is  a  system  of  forces  in  a  plane  (a  "  uniplanar" 
system  of  forces). 

18.  Contact  Forces  or  Pressures. — If  one  of  the  two  bodies  in 
contact  is  rounded  at  the  point  of  contact,  while  the  other  is  quite 
flat  at  that  point,  the  action-line  of  their  mutual  pressure  neces- 
sarily lies   in    a  perpendicular,   or  normal,  to   the   latter  flat 
surface,  and  passes  through  the  point  of  contact.     Hence,  the 
shapes  of  the  bodies  being  known,  this  action-line  becomes  known 
on  inspection.     Let  this  be  called  "  flat-contact  pressure"     (N.B. 
As  a  better  definition  of  flat-contact  pressure,  we  might  describe 
it  as  a  pressure  occurring  in  such  a  way  that  its  action-line  can- 
not be  materially  changed  by  any  slight  motion  of  one  piece  rela- 
tively to  the  other  during  the  small  alteration  of  form  and  posi- 
tion which  actually  takes  place  when  a  load  is  gradually  placed 
on  the  structure.} 

But  if  the  mode  of  connection  of  the  two  bodies  is  &  pin- 
connection,  that  is,  if  one  body  carries  a  round  pin  or  bolt  fitting 
(somewhat  loosely)  in  a  corresponding  ring  forming  part  of  the 
other  body,  we  are  unable  to  say  in  advance  just  where,  on  the 
inner  circumference  of  the  ring,  the  contact  (and  accompanying 
pressure)  is  going  to  be.  Wherever  the  point  is,  all  that  we  can 
immediately  say  as  to  the  action-line  of  the  force  is  that  it  passes 
through  the  centre  of  the  circle,  its  direction  (if  determinate  at  all) 
being  found  from  a  consideration  of  the  other  forces  acting  on 
the  piece  in  question.  This  will  be  illustrated  later. 

Such  a  pressure  may  be  called  a  hinge-pressure. 

19.  Classification  of  Rigid  Bodies  under  Uniplanar  Systems  of 
Forces. — As  conducive  to  clearness   in    subsequent    matter,  the 
rigid  bodies  composing  a  structure  will  be  named  according  to 
the  number  of  forces  acting  on  each  ;  and  these  forces  consist  of 
gravity  actions  (i.e.,  weights)  and  of  the  pressures  exerted  by 
neighboring  pieces  on  the  piece  in  question.    The  resultant  gravity 
action  on  a  single  piece,  or  member  of  the  structure,  is  a  single 
force,  called  its  weight,  acting  vertically  downward  through  its 


18 


NOTES   AND   EXAMPLES   IN   MECHANICS. 


FIG.  17. 


centre  of  gravity.  (Of  course,  the  action  of  gravity  is  distributed 
over  all  the  particles  of  a  body,  but  the  above-mentioned  single 
force  is  the  full  equivalent  of  these  distributed  forces  as  far  as 
the  equilibrium  of  the  piece  is  concerned;  though  not  such  as  re- 
gards the  straining  action  on  the  piece.  With  these  straining 
actions  we  cannot  deal  here ;  they  will  be  treated  later  in  the 
proper  place.  In  many  cases  the  whole  weight  of  a  piece  is  so 
small  in  comparison  with  any  of  the  other  forces  of  the  system 
acting  on  that  piece  that  no  appreciable  error  is  made  in  regard- 
ing it  as  without  weight.  JSTotice  is  always  given  in  such  cases.) 
20.  "  Two-force  Pieces  "  and  their  Treatment. — A  "  two-force" 
piece  being  a  piece  on  which  only  two  forces  act,  if  the  weight 
of  the  piece  is  considered  there  is  but  one  other 
force.  For  example,  Fig.  17,  a  body  of  weight 
G  hangs  by  a  stem  and  ring  which  form  a  rigid 
part  of  it,  on  a  pin  projecting  from  a  fixed  sup- 
port. 

Since,  evidently,  the  equilibrium  of  a  two- 
force  piece  requires  that  the  two  forces  shall  be 
equal  and  opposite  and  act  in  the  same  line,  the  piece  A  .  .  B 
will  not  be  in  equilibrium  unless  the  centre  of  gravity  c  lies  in  a 
vertical  line  drawn  through  the  centre  of  the  circular  section  of 
the  pin  at  A.  Here  we  have  an  instance  of  the  final  full  deter- 
mination (by  the  necessity  of  its  being  vertical)  of  the  action- 
line  of  a  hinge-pressure,  concerning  which  we  know  in  advance, 
only  that  it  passes  through  the  centre  of  the  circle  at  A.  We 
find,  therefore,  that  the  pressure  of  the  pin  at 
A  against  the  ring  of  the  rigid  body  A  .  .B, 
hanging  at  rest,  must  have  a  direction  verti- 
cally upward,  and  an  amount,  V,  numerically 
equal  to  G,  while  its  action-line,  c  .  .  d,  passes 
vertically  through  the  centre  of  the  hinge-circle. 
In  Fig.  18  is  another  two-force  piece,  in 
which,  for  equilibrium,  the  same  result  is 
reached,  but,  the  stem  being  curved,  the  straining  action  in  it 
is  of  a  bending  nature ;  whereas,  in  Fig.  17  it  is  a  simple  tension, 
or  stretching  action. 

The  case  of  a  two-force  piece  whose  weight  is  neglected  is 


FIG.  18. 


THREE-FORCE   PIECES. 


19 


Sett  Crank 


-G 


FIG.  19. 


instanced  in  Fig.  19,  where  the  two-force  piece  A  .  .  B  is  sub- 
jected to  tension  through  the 
action  of  a  suspended  weight,  G, 
and  a  bell-crank  lever,  BCD. 
The  hinge-pressure  at  A,  from 
the  support  8  against  the  piece 
A  .  .  B,  must  pass  through  the 
centre  of  the  circle  at  A  ;  while 
also  the  hinge-pressure  at  B,  of 
the  bell-crank  against  the  piece 
A  .  .  B,  must  pass  through  the 
centre  of  the  hinge-circle  at  B. 
But  these  two  hinge-pressures  are  the  only  forces  acting  on  the 
piece  A  .  .  B,  and  for  the  equilibrium  of  a  two-force  piece 
must  be  equal,  opposite,  and  coincident  as  to  action-line ;  that 
is,  these  pressures  must  both  act  in  the  line  joining  the  centres 
of  the  two  circles.  A  B'  shows  this  piece  represented  as  a  free 
body,  with  the  equal  and  directly  opposite  forces  Pr  and  P'  act- 
ing at  the  ends.  As  to  the  value,  or  amount,  of  this  force  Pr,  it 
cannot  be  found  until  the  case  of  the  bell-crank  has  been  treated, 
depending,  as  it  does,  not  only  upon  the  design  of  the  bell-crank 
and  the  amount  of  the  load  #,  but  also  upon  the  position  of  the 
piece  AB  itself. 

21.  Three-force  Pieces. — If   a  rigid  body  is  at  rest  (i.e.,  in 
equilibrium)  under  the  actions  of  three  forces,  it  is  evident  that 
these  forces  must  have  action-lines  intersecting  in   a  common 
point,  and  that  each  force  must  be  the  equal  and  opposite  of  the 
diagonal  of  the  parallelogram  formed  on  the  other  two  as  sides 
(laid  off  to  scale)  ;  for  any  one  of  them  'must  ~be  the  anti-resultant 
of  the  other  two.     (See  §  15,  M.  of  E.)     [In  the  particular  case 
where  the  forces  are  parallel  the  intersection-point  is  at  infinity 
and  the  value  of  any  one  of  the  forces  is  numerically  equal  to  the 
sum  of  the  other  two  (algebraic  sum).] 

22.  Example  of  a  Three-force  Piece. — The  bell-crank  BCD  of 
Fig.  19  furnishes  an   instance.     This  body  is  subjected  to  the 
three  hinge-pressures  at  B,  C,  and  D,  respectively.     That  at  D 
is  a  vertical  downward  pressure  6r',  equal  to  the  weight  G  of  the 


20  NOTES  AND   EXAMPLES   IN   MECHANICS. 

two-force  piece  DE.  The  action-line  of  the  hinge-pressure  at  B 
has  been  found  by  the  previous  consideration  of  the  two-force 
piece  AB,  and  is  a  .  .  b.  The  hinge-pressure  P' ',  at  C\  passes 
through  the  centre  of  the  corresponding  circle,  but  its  action-line 
is  as  yet  unknown.  The  problem,  then,  stands  thus :  Of  the 
three  forces,  G',  P\  and  P"  ^  under  whose  action  the  bell-crank 
is  in  equilibrium,  G'  is  known  in  amount  and  line  of  action,  P' 
is  known  as  to  action-line  but  not  in  amount,  while  as  to  P"  we 
know  neither  its  amount  nor  its  direction  but  simply  one  point, 
6",  of  its  action-line.  However,  since  the  three  action-lines  must 
meet  in  a  common  point,  we  need  only  note  the  intersection,  0, 
of  the  known  action-lines  of  P1  and  G'9  and  join  o  with  C 
(centre),  in  order  to  determine  C .  .  o,  the  action-line  of  P" . 
Next,  as  to  finding  the  amounts  of  both  P'  and  P" ,  consider  that 
P"  is  the  anti-resultant  of  P'  and  G',  and  that  therefore  the 
(ideal)  resultant  of  P'  and  G'  must  act  along  o  .  .  C;  hence 
lay  off  o  .  .  m  by  scale  to  represent  G'  and  through  m  draw  a 
line  ||  to  o  .  .  A  intersecting  o  .  .  C  in  some  point  r,  and  draw 
r  .  .  k  ||  to  G',  to  determine  ~k  on  the  line  o  .  .  B.  Then  o  .  .  n, 
laid  off  along  O .  .  o,  and  =  o  .  .  r,  but  in  the  opposite  direction 
from  0,  gives  the  amount  and  direction  of  P".  For  o  .  .  r  is 
the  resultant  of  P'  and  G' ,  and  o  .  .  n  is  its  equal  and  opposite. 

Of  course  P'  and  P"  must  be  measured  by  the  same  force- 
scale  that  was  used  in  laying  off  o  .  .  m  =  G' . 

We  can  see  by  inspection  of  the  figure  that  if  the  position  of 
the  link,  or  two-force  piece,  AB,  were  changed  in  such  a  manner 
that,  while  the  line  A  .  .  B  continues  to  pass  through  0,  the  pin- 
joint,  or  hinge,  B  is  caused  to  approach  nearer  and  nearer  to  C9 
the  forces  P"  (always  equal  to  the  ideal  o  .  .  r)  and  P'  both 
increase  without  limit ;  for  the  point  r  moves  out  from  0,  m 
being  fixed  (i.e.,  the  load  G  remains  invariable)  until,  when  B  is 
infinitesimally  near  to  C,  m  .  .  r  is  ||  to  o  .  .  C  and  r  is  at 
infinity. 

The  method  just  pursued  is  a  graphic  one ;  analytically,  we 
would  proceed  thus  :  since  the  system  of  forces  G',  Pr,  and  P" 
is  balanced,  i.e.,  in  equilibrium,  the  algebraic  sum  of  their 
moments  about  any  point  in  the  plane  must  vanish,  i.e.,  =  0. 


THREE-FORCE   PIECES  —  ANALYTIC   TREATMENT.  21 

(See  §  16.)  Take  an  origin,  or  centre  of  moments,  at  C  and 
denote  by  d  and  c  the  lengths  of  the  perpendiculars  let  fall  from 
C  upon  the  action-lines  of  G'  and  P  ',  respectively.  (These 
lengths  may  be  obtained  trigonometrically  from  given  distances 
and  angles,  but  are  most  easily,  and  with  sufficient  precision, 
scaled  off  from  an  accurate  drawing.)  With  C  as  origin  the 
lever-arm  of  the  force  P"  is  zero  and  hence  the  moment  of  this 
force  is  zero  ;  consequently  this  force  does  not  enter  the  moment- 
equation,  which  therefore  will  contain  but  one  unknown  quantity, 
viz.,  the  amount  of  the  force  P1  '. 

The  resulting  moment-equation  (following  the  routine  recom- 
mended at  the  foot  of  page  33,  M.  of  E.)  is 

P'  .G-  Gf.d  +  P"x  0  =  0.  .(1);    whence  P'  =  -G'  .  .  .  (1) 

G 

becomes  known  ;  and  since  P"  is  the  anti-resultant  of  P'  and 
G-',  we  have  also,  a  being  the  angle  between  the  action-lines  of 
the  latter  forces, 


P"  =  ^Pj^^7^^PjW~^~a  .....     (2) 
(See  p.  7,  M.  of  E.) 

23.  Other    Examples    of    Three-force    Pieces.  —  The    ordinary 
straight  lever,  with  flat-contact  supports,  is  shown  in  Fig.  20. 
Since  the  pressures  (or  reactions]  of  the 
supports  against  the  lever  must  be  ~| 
to   the   axis  of  the   latter,  and   hence 
parallel,  in  this  case  the  action-line  of 
the  third  force  P'  must  be  made  "|  to 
the  lever.     Otherwise  equilibrium  could  FlG-  2°- 

not  he  maintained,  for  the  point  of  intersection  of  the  three  force- 
lines  is  fixed  by  the  intersection  of  the  fiat-contact  pressures 
P"  and  Q  ;  at  infinity  in  this  instance. 

Given  P',  we  determine  Q  and  P"  by  considering  the  lever 
as  a  free  body  under  a  system  of  three  forces  in  equilibrium  (in 
a  plane),  taking  a  moment-centre  at  B  so  as  to  exclude  the 
unknown  force  P"  from  the  equation  ;  and  obtain  first,  as  a 
moment  equation, 

Qb-P'a  =  0,    .   .    or,    Q  =     Pt' 


22  NOTES   AND   EXAMPLES   IN  MECHANICS. 

and  then,  by  summing  all  the  components  of  the  three  forces  in  a 
direction  at  right  angles  to  the  lever, 

+  P"-0-P'  =  0;     whence    P" '  =  Q  +  P1 ; 

and  thus  the  two  unknown  forces  have  been  found  in  amount 
(and  are  already  known  in  position). 

Obviously,  if  Q  is  given,  being,  e.g.,  the  weight  of  a  body  to 
be  sustained,  we  compute  in  a  similar  manner  the  necessary  force 
P'  to  be  applied  at  C  and  the  resulting  pressure,  P"  =  P'  +  ft 
at  the  support  or  fulcrum  B. 

It  is  also  evident  that  the  smaller  the  distance  1}  is  made  in 
comparison  with  a,  the  greater  the  pressures  P"  and  ft  for  a 
given  P1 ';  in  fact,  as  5  approaches  zero,  P"  and  Q  increase  with- 
out limit.  For  a  given  Q  and  diminishing  value  of  the  ratio 
5  :  a,  the  necessary  force  P1  decreases  toward  zero.  (N.B.  In 
these  cases  the  weight  of  the  lever  itself  is  neglected.) 

As  showing  how  the  possibility  of  equilibrium  may  be  de- 
pendent in  some  cases  on  the  design  and  position  of  the  support- 
ing surfaces,  let  us  consider  the  curved  lever 
in  Fig.  21,  where  the  supporting  surfaces  A 
and  B  are  capable  of  furnishing  only  flat- 
contact  pressures  or  reactions,  whose  direc- 
tions, A  .  .  O  and  B  .  .  O,  are  fixed,  being 
normal  to  the  respective  smooth  and  flat 
surfaces  of  contact.  (N.B.  Smooth  sur- 
faces are  postulated  in  all  the  present  prob- 
lems ;  rough  surfaces  will  be  considered 
later.) 

The  intersection,  O,  of  their  action- lines  is  therefore  fixed, 
and  if  a  force  P'  is  to  be  applied  at  a  given  point  C  to  induce 
pressures  at  A  and  B,  its  line  of  action  must  be  taken  along 
C .  .  O  ;  otherwise  the  lever  will  begin  to  move  out  of  its  pres- 
ent position  (weight  of  lever  neglected).  Given,  then,  the  force 
P'  along  C  .  .  O,  we  determine  P"  and  Q  for  equilibrium,  in  the 
same  manner  as  before  shown,  by  filling  out  the  parallelogram 
O  .  m  .  r  .  Ic,  in  Fig.  21,  precisely  as  was  done  in  Fig.  19  (except 
that  the  P'  of  this  problem  corresponds  to  the  Gr  of  Fig.  19). 


CURVED   LEVER.       REDUNDANT   SUPPORT. 


FIG.  22. 

A  we  have  an  un- 
action-line  A  .  .  O. 


However,  if  we  assume  any  direction  at  pleasure  for  the 
action-line  of  P'  through  the  point  (7,  and  wish  to  secure  equilib- 
rium, we  have  only  to  change  the  mode  of 
support  at  A  or  at  B  (say  B\  into  a  pin- 
joint  or  hinge-support ;  for  the  direction  of 
a  hinge-pressure  is  not  determined  solely 
by  the  nature  of  this  mode  of  support ;  the 
only  restriction  upon  it  known  in  advance  is 
that  it  must  pass  through  the  centre  of  the 
hinge-circle,  its  direction  being  determined 
by  other  relations.  This  change  being  made, 
we  have  Fig.  22,  in  which  P'  is  given,  or 
assumed,  both  in  amount  and  position.  At 
known  flat-contact  pressure  Q  in  a  known 
The  intersection  O  of  the  two  action-lines  A  .  .  0  and  O .  .  0  must 
be  a  point  in  the  action-line  of  P" ,  the  unknown  hinge-pressure  ; 
i.e.,  B  .  .  0  is  the  action-line  of  the  latter,  while  its  amount,  as 
well  as  that  of  Q,  is  found  by  a  construction  like  that  in  Fig.  19, 
which  need  not  be  explained  again.  The  results  are  that 

P"  =  0  .  .  ft,  and  Q  =  0  .  .  n  (the  equal  and  opposite  of  0 .  .  r\ 

these  lines,  or  rather  lengths,  representing  forces  on  the  same 
scale  as  that  on  which  0  .  .  m  represents  the  first  given  force  Pf . 
24.  Redundant  Support.— If  the  two  supports,  A  and  B,  of 
the  lever  are  l>oth  hinge-joints,  as  shown  in  Fig.  23,  the  body  or 
lever  ABC  is  redundantly  supported  ;  for 
now  the   hinge-pressures  at  A  and  B  are 
indeterminate,  from  simple    Statics  alone, 
but  depend  on  the  form  and  elasticity  of  the 
lever  itself  and  upon  the  degree  of  loose- 
ness of  fitting  of  the  hinge-rings  around  the 
pins  of  the  supports  A  and  E,  and  upon 
any  slight  elastic  yielding  of  the  latter ;  as 
well  as  upon  the  amount  and  position  of 
FIG'23'  theforceP'. 

In  fact,  if  the  body  is  elastic,  and  we  have  to  "  spring"  it  to 


NOTES  AND  EXAMPLES  IN  MECHANICS. 


cause  the  rings  to  fit  over  the  pins,  pressures  are  produced  at  the 
supports  before  the  application  of  any  force  at  C. 

From  simple  Statics,  then,  all  that  we  can  claim  is  that  the 
action-lines  of  the  hinge-pressures  P"  and  Q  must  intersect  in 
some  point  0  situated  on  the  given  action-line  of  the  given  force 
P' ;  but  have  no  means  of  fixing  the  position  of  this  point  0., 
Problems  of  this  nature,  therefore,  cannot  be  treated  until  the 
theory  of  Elasticity  is  presented,  and  then,  as  will  be  seen,  only 
in  comparatively  simple  cases.  In  attempting  an  analytical  treat- 
ment in  this  case  we  should  find  that  it  presents  four  unknown 
quantities ;  whereas  from  simple  Statics  only  three  equations 
(independent  equations)  can  be  obtained  for  the  equilibrium  of  a 
system  of  forces  in  a  plane ;  hence  the  indetermi nation. 

25.  Four-force  Pieces. — If  the  rigid  body  is  in  equilibrium 
under  a  system  of  four  forces  one  of  which  is  given  both  in 
amount  and  position,  while  the  action-lines  of  all  the  other  three 
are  known,  the  amounts  of  those  three  can  be  determined. 

A  simple  graphic  method  for  solving  this  case  is  based  on 
the  obvious  principle  that  if  four  forces  are  in  equilibrium  the 
(ideal)  resultant  of  any  two  must  be  equal  and  opposite  to  the 
resultant  of  the  other  two  and  have  the  same  action-line. 

26.  The  Simple  Crane. — A  convenient  example  of  a  four-force 
piece  is  presented  by  the  simple  kind  of  crane,  ABC,  in  Fig.  24, 

consisting  of  a  single  rigid  body,  of 
curved  form.  Its  lower  extremity  rests 
in  a  shallow  socket,  while  at  B  the  edge 
of  the  (wharf)  floor  furnishes  lateral 
support.  We  neglect  the  weight  of  the 
crane,  and  assume  that  no  pressures  are 
induced  -at  A  and  B  unless  the  crane 
bears  a  load  ;  i.e.,  that  the  parts  are 
FIG.  24.  loosely-fitting  at  A  and  B.  Placing 

now  a  known  load  at  (7,  viz.,  Pl ,  we  note  that  in  preventing  the 
overturning  of  the  crane  the  right-hand  edge  of  the  floor  at  B 
reacts  against  the  crane  with  some  horizontal  pressure  P^ 
(horizontal,  since  the  surface  of  contact  is  vertical),  while  at  A 
there  are  two  surfaces  under  pressure,  one  of  which  is  horizontal, 


Simple  Crane 


SIMPLE   CRANE — GEAPHICALLY   AND   ANALYTICALLY.    25 

while  the  other  is  the  left-hand  vertical  side  of  the  socket ;  there- 
fore at  A  we  have  the  vertical  and  horizontal  reactions,  Pt  and 
P3,  both  unknown  in  amount. 

Now  pair  off  the  four  forces  at  convenience  ;  for  example,  as 
in  our  figure,  pair  off  Pl  with  P% ,  noting  the  intersection,  0,  of 
their  action-lines  ;  Ps  and  jP4  constitute  the  other  pair  and  inter- 
sect at  o' . 

Since  o  is  a  point  in  the  action-line  of  the  resultant  of  JP,  and 
PI,  while  or  is  a  point  in  that  of  the  resultant  of  Pt  and  P4 ; 
and  since  these  two  resultants  must  have  a  common  action-line 
for  equilibrium,  that  common  action-line  must  be  o  .  .  o'.  Pro- 
ceed therefore  as  follows :  Prolong  G .  .  o  and  make  o  .  .  m  equal 
to  PI  by  any  convenient  scale.  We  know  that  the  diagonal 
of  the  parallelogram  formed  on  JP,  and  jPa  must  lie  on  the  line 
o  .  .  of  (prolonged,  here).  This  parallelogram  is  found  by  making 
m  .  .  r  ||  to  Pz  to  determine  r  on  o  .  .  o' ;  then  drawing  r  .  .  k  \\  to 
m  .  .  o  to  intersect  A  .  .  o  in  some  point  k.  Now  prolong  o  .  .  o' 
beyond  </,  making  of  .  .  r'  equal  to  o  .  .  r.  Then  of  .  .  r'  is  the 
resultant  of  the  unknown  Pt  and  jP4,  and  by  drawing  the  proper 
parallels,  as  shown,  we  resolve  o'  .  .  r'  in  the  directions  of  those 
forces  and  thus  determine  o'  .  .  mf  =  P3,  and  o'  .  .  k'  =  P< ; 
o  .  .  ~k,  already  found,  is  equal  to  P^ ;  all,  of  course,  on  the  as- 
sumed scale  of  force  (which  scale  is  entirely  independent  of  the 
scale  for  distances  used  in  laying  out  the  dimensions  of  the  crane 
on  the  paper). 

Evidently,  in  this  particular  problem,  Pl  =  P8,  and  P2  =  P4 . 
(N.B.  It  will  be  noted  that  by  this  method  the  directions  of 
pointing  of  the  unknown  forces  are  found,  as  well  as  their  magni- 
tudes.) It  is  also  easily  noted,  on  inspection,  that  if  the  distance 
B  .  .  A  is  made  shorter  and  shorter  and  diminishes  towards 
zero,  Pl  remaining  the  same  in  amount  and  position,  the  forces 
P^  and  PI  increase  without  limit  (i.e.,  become  infinite  for 
B  .  .  A  =  0). 

27.  Simple  Crane  Treated  Analytically.  Single  Rigid  Body. — 
Take  A  as  a  centre  of  moments  (for  by  so  doing  we  exclude  the 
two  unknown  quantities  P^  and  jP3  from  the  moment-equation, 
since  then  their  lever-arms  are  both  zero  ;  and  thus  obtain  an 


26  NOTES   AND   EXAMPLES   IN   MECHANICS. 

equation  containing  only  one  unknown  quantity).  The  lever- 
arm  of  Pt  is  Ao',  and  that  of  P1 ,  Ao.  Hence 

IT 
+  P. .  Ao'  —  Pl .  Ao  +  0  +  0  =  0  ;  whence  P4  =  —  .  Pt.  .  (1) 

jy 

The  balancing  of  horizontal  components  gives 

'  An 

+  P,  -  P.  =  0,   whence    P2  =  P4 ;    i.e.,    P2  =  S^ .  Pl ,  .  (2) 

J.0' 

Also,  from  the  balancing  of  vertical  components, 

P,-Pl  =  0',le.)P3=P1 (3) 

28.  Multi-force  Pieces. — Since  we  can  at  most  determine  the 
amounts  of   only  three  unknown  forces  (action-lines  given)   in 
a  uniplanar  system  in  equilibrium ;  and  since  to  determine  any 
forces  at  all  there  must  be  one  force  given  in  all  of  its  elements 
(i.e.,  amount  and  position) ;  it  follows  that  when  the  system  con- 
sists of  more  than  four  forces,  all  but  three  of  them  must  be  given 
in  amount  and  position,  and  also  the  action-lines  of  the  unknown 
three  must  be  given,  if  the  problem  is  to  be  a  determinate  one. 
In  such  a  case  it  is  a  simple  matter  to  combine  the  known  forces 
into  a  single  resultant  by  successive  applications  of  the  parallelo- 
gram of  forces,  and  thus  reduce  the  problem  to  that  of  a  four- 
force  piece,  treating  it  then  as  in  the  last  figure  (Fig.  24),  the 
resultant  of  all  the  known  forces  playing  the  part  of  P3  in  that 
figure.     (N.B.  To  find  graphically  the  resultant  of  two  parallel 
forces  we  can  use  a  construction  like  that  in  Fig.  10  or  11  of  pp. 
13  and  14,  M.  of  E.) 

29.  Compound  Crane  on  Platform-car. — As  an  example  of  the 
itility  of  the  foregoing  principles,  let  us  apply  them  to  the  case  of 

the  several  rigid  bodies  constituting  the  composite  (or  built-up) 
crane  of  Fig.  25. 

Here  we  have  an  assemblage  of  seven  rigid  bodies,  forming 
a  rigid  structure  at  rest ;  viz.,  the  jib,  ABC;  the  tie-rod^  DB ; 
the  mast,  UADE;  the  tie-rod,  FE\  the  platform,  FS',  and  the 
two  wheel-pairs,  M  and  R.  (Each  pair  of  wheels  and  its  axle 
form  together  a  single  rigid  body.)  The  track  is  level. 

For  simplicity  we  shall  consider  the  platform  alone  as  having 
weight,  viz.,  a  force  Gl ,  applied  in  the  centre  of  gravity,  as 


COMPOUND    CRANE   ON    CAR — WHEEL-PRESSURES.         27 

shown ;  while  the  extremity  C  of  the  jib  is  to  carry  a  load  G. 
We  have  given,  therefore,  the  two  forces  Gl  and  G,  and  all 
angles  and  distances  concerned,  and  are  required  to  find  the 
pressures  induced  at  the  various  points  of  contact  between  the 


Compound  Crane. 


FIG.  25. 


parts  of  the  structure,  and  under  the  wheels.  (The  final  practical 
object  is,  of  course,  to  give  sufficient  strength  to  the  parts  for 
their  respective  duties,  a  matter,  however,  which  cannot  be 
entered  into  here,  belonging,  as  it  does,  to  the  topic  of  "  Mechan- 
ics of  Materials.")  Only  analytical  methods  will  be  used. 

29a.  The  Wheel-pressures.  —  Let  us  first  consider  the  whole 
structure  as  a  free  body.  The  only  forces  external  to  this  free 
body  are  the  two  gravity  forces  G  and  Gl  ,  and  the  vertical  up- 
ward pressures,  V0  and  Vn  ,  of  the  rails  against  the  wheels  ;  these 
constitute  a  system  of  parallel  forces  in  equilibrium  and  are  all 
shown  in  Fig.  25.  (NOTE.  —  The  pressures  between  any  two  con- 
tiguous parts  of  our  present  free  body  do  not  need  to  appear  in 
this  system  for  the  reason  that,  if  introduced,  they  would  form  a 
balanced  system  among  themselves  and  might  therefore  be  re- 
moved without  affecting  equilibrium.  For  example,  at  D  the 
ring  of  the  tie-rod  presses  horizontally  and  to  the  right  against 
the  pin  of  the  mast  ;  and  the  pin,  from  the  principle  of  action 
and  reaction,  presses  the  ring  with  an  equal  horizontal  pressure 


28  NOTES   AND   EXAMPLES   IN   MECHANICS. 

toward  the  left ;  but  both  the  ring  and  the  pin  belong  to  the 
free  body  now  under  consideration  in  Fig.  25,  and  if  one  of 
these  pressures  is  inserted  in  the  system  the  other  must  be  placed 
there  with  equal  right,  and  the  two  then  annul  each  other's  influ- 
ence in  all  the  equations  of  equilibrium.  We  therefore  conclude, 
in  general,  that  the  mutual  actions  Between  the  parts  of  a  given, 
free  body  in  equilibrium  may  be  omitted  in  applying  the  con- 
ditions of  equilibrium?) 

With  0  as  a  centre  of  moments,  then,  we  have  for  the  free 
body  of  Fig.  25  (see  figure  for  notation) 

TT  \       ^  -rr       G(a-\-  m)-f-  G.a. 

Vnan  -  G(a  +  m)  -  G&  =  0  ;       .-.  Vn  =  -        —A.         _> ; 

while  by  summing  vertical  components 

V0  +  Vn-  G-  £,  =  0;  hence  V0=G+G1-Vn. 
29b.  Pressures  at  the  Joints ;  D,  B,  and  A  (Fig.  25). — By  in- 
spection we  see  that  the  tie-rod  DB  is  a  two-force  piece  (its  own 
weight  neglected) ;  that  is,  the  hinge-pressures  at  D  and  B  must 
have  a  common  action-line,  viz.,  D  .  .  B.  DB  is  a  straight  two- 
force  piece  •  or  "straight  lin~k"  as  we  shall  hereafter  call  it ;  and 
is  subjected  to  a  tensile  action  along  its  axis  (tension,  here ;  as  we 
note  by  inspection  ;  a  straight  link  under  compressive  action  is 
called  a  strut,  or  compression-member). 

The  Jib  as  a  free  Body.     Fig.  26.     It  is  a  three-force  piece, 
being  acted  on  by  the  known  vertical  down- 
a       v.^  ward  force  G  at  67,  by  an  unknown  horizontal 
force  T  (directed  toward  the  left)  at  B  (T 
._  being  the  pull  of  the  tie-rod),  and  an  unknown 
hinge-pressure  P  at  A,  making  an  unknown 
angle  a  with  the  horiozntal. 

NOTE. — Since  the  mast  which  acts  on  the 
jib  at  A  is  not  a  two-force  piece,  we  have  no 
means  of  knowing  the  position  of  the  action- 
line  of  P  from  a  mere  inspection  of  the  mast, 
as  we  did  with  the  tie  DB  and  the  hinge-pressure  at  B.  Of 
course,  graphically,  P  passes  through  the  point  m  where  the 
action-lines  of  the  other  two  forces  intersect ;  but  as  we  are  now 


TENSION  IN  THE  TIE-KOD.  29 

using  analytical  methods,  we  shall  replace  P,  which  is  unknown 
in  amount  and  in  position,  by  its  (ideal)  horizontal  and  vertical 
components,  Ph  and  Pv  (i.e.,  by  two  unknown  forces  in  known 
action -lines).  We  thus  have  a  four-force  piece  to  deal  with. 

If  we  take  A  as  a  centre  of  moments,  the  force  T  will  be  the 
only  unknown  quantity  in  the  corresponding  moment-equation, 
which  is  Tl}  —  Got  —  0 ;  whence  we  have,  for  the  tension  T  in 
the  tie-rod,  T=  G(a  ~-  1).  (This  T  is  the  value  of  the  hinge- 
pressure  at  D  and  also  that  at  B,  in  Fig.  25.) 

Assuming  an  axis  X  horizontal  and  _F  vertical,  we  now  have 

from    2X=  0,      Ph  -  T=  0  ;      or,    Ph  =  T\      while  from 

2Y=0,      PV-G=0',      or,    Pv  =  G ;   and  we  now 

p 

easily  find  P  itself,  since  P  =   VPJ  +  P;  ;  while  tan  a  =  ^~. 

•** 

29c.  Tension  in  the  Tie-rod  FE.  As  with  DB,  so  with  FE 
we  note  by  inspection  that  it  is  a  two-force 
piece,  so  that  the  hinge-pressure,  T ',  at  E, 
Fig.  25,  though  unknown  in  amount  as  yet, 
must  have  E . .  jFas  action-line.  The  force 
Tf  and  the  pressures  (or  supporting  forces) 
II  and  V  exerted  by  the  left-hand  side  and 
bottom,  respectively,  of  the  shallow  socket 
U  against  the  foot  of  the  mast,  are  the  three 
unknown  forces  in  known  action-lines  acting 
on  the  free  body  shown  in  Fig.  27,  consist- 
ing of  the  jib,  mast,  and  tie-rod  DB.  The 
mutual  actions  between  these  three  bodies  are  internal  to  the 
free  body  taken  and  are  hence  omitted  (see  note  in  §  29a),  the 
external  system  consisting  of  Tr,  II,  V,  and  G\  all  in  known 
action-lines,  G  being  the  only  force  known  in  amount.  By 
moments  about  the  point  £7,  we  have 

T'n  -  Ga  =  0 ;    whence  T'  =  -  .  G.     .    .     .    (1) 

From  ^X=  0,  IT-  T'  cos  a'  =  0,  and  .  • .  JI=  -  G  cos  a'.  (2) 

n 

=  0,  F-^-r'sinar'^O;  .-.  V=  G+  T  sm  a'.  (3) 


30 


NOTES   AND   EXAMPLES   IN   MECHANICS. 


FIG.  28. 


It  will  be  noticed  that  we  have  not  made  use  of  the  mast  as  a 
separate  free  body.  This  might  have  been  done  as  a  means  of 
finding  the  three  forces  just  determined,  T ',  H,  and  Y',  since  the 
hinge-pressures  at  D  and  A  had  already  been  de- 
duced ;  but  the  process  would  have  been  more 
roundabout. 

However,  as  a  reminder  of  the  principle  of  T 
action  and  reaction  and  of  the  definition  of  force, 
Fig.  28  is  presented,  showing  the  system  of  forces 
we  should  have  to  deal  with  in  treating  the  mast 
as  a  free  body  ;  and  also  Fig.  29,  representing  the 
car-platform  and  the  two  pairs  of  wheels  as  a  single 
free  body,  with  the  external  forces  acting.     The 
student  will  note  that  the  H  and  V  in  Fig.  28  are 
the  equals  and  opposites,  respectively,  of  those  in  Fig.  29.     A 
similar   statement   may  be  made  for  the  T'  of  those  figures. 

Again,  the  P  and  T  of  Fig. 
s  28  are  the  equals  and  oppo- 
sites of  the  P  and  T  of  Fig. 
26  (action  and  reaction).  The 
plane  of  the  crane  being  sup- 
posed to  be  midway  between  the  two  wheels  of  each  pair,  the 
pressure  V0  is  equally  divided  between  the  two  wheels  forming 
the  pair  on  the  left.  Similarly  at  JV,  with  Vn . 

30.  Simple  Roof  and  Bridge  Trusses ;  Hitter's  Method  of  Sections. 
— A  truss  is  an  assemblage  of  straight  pieces  jointed  together  in 
one  plane.  If  the  joints  consist  of  pins  (one  in  each  joint) 
inserted  through  holes  or  rings  in  the  ends  of  the  pieces  (or 
u  members")  the  truss  is  said  to  be  "pin-connected";  while  if  the 
ends  of  the  pieces  meeting  at  each  joint  are  rigidly  riveted 
together  (a  favorite  method  in  Europe)  the  truss  is  said  to  have 
riveted  connection. 

In  the  first  case,  pin-connection,  each  piece  is  free  to  turn 
about  the  pin,  independently  of  all  other  pieces,  during  the 
gradual,  though  slight,  change  of  form  which  the  truss  undergoes 
in  the  gradual  settling  of  a  load  upon  it,  and  the  stresses  induced 
in  the  pieces  are  called  "  primary  stresses"  (whereas,  with  riveted 


FIG.  29. 


RITTER'S  METHOD  OF  SECTIONS — EXAMPLE  OF  TRUSS.  31 


joints,  other,  and  additional,  stresses,  called  "  secondary  stresses," 
are  caused  in  the  pieces,  from  the  constraint  exerted  on  each 
other  by  the  members  meeting  at  each  joint). 

Confining  ourselves  to  the  consideration  of  pin-connected 
trusses,  constructed  so  that  each  piece  connects  no  more  than  two 
joints,  and  loaded  only  at  the  joints  (its  own  weight  being  con- 
sidered as  concentrated  at  the  various  joints),  we  note  that  in 
such  a  case  each  member  must  be  a  straight  two-force  piece  ^  or 
straight  link  (neglecting  its  own  weight) ;  i.e.,  it  is  subjected  to  a 
simple  tension  or  com  pression  (according  as  it  is  acting  as  a  tie 
or  a  strut)  along  its  axis. 

NOTE. — If  such  a  straight  link  be  conceived  divided  into  two 
parts  (for  separate  treatment)  by  any  imaginary  transverse  plane 
or  surface  passing  between  the  joints  connected  by  that  piece, 
the  action  or  force  exerted  on  one  of  these  parts  by  the  other  is 
a  pull  (if  tension)  or  thrust  (if  compression),  applied  at  the  section 
and  directed  along  the  axis  or  central  line  of  the  piece.  For 
example,  in  the  truss  of  Fig.  30,  pin-connected,  and  composed  of 


t 


"  straight  links,"  if  we  wish  to  consider  free  the  portion  on  the 
left  of  the  imaginary  cutting  surface  A  . .  B,  the  system  of  forces 
acting  on  the  ideal  body  so  obtained  (see  Fig.  31)  consists  of  the 
abutment  reaction  F"0 ,  the  loads  G,  G' ,  and  G"  (at  the  joints 
#,  &,  and  <?),  and  the  three  forces  (pulls  or  thrusts)  P,  Q,  and  T, 
acting  at  the  (cut)  ends  of  the  three  pieces  intersected  by  this 
surface  A  . .  B. 

If  the  pier  reactions  F0  and  Vn  have  been  already  determined 
by  a  consideration  of  the  whole  truss  as  a  free  body,  F0  is  now  a 
known  quantity,  and  we  may  go  on  to  find  the  values  of  the 


82  NOTES   AND   EXAMPLES   IN   MECHANICS. 

three  stresses  (i.e.,  pulls  or  thrusts)  P,  Q,  and  T,  by  methods 
already  illustrated.  It  is  evident  from  inspection  that  the  force 
or  stress  in  the  lower  horizontal  piece  is  a  pull  (P) ;  and  that  the 
stress  (T)  in  the  upper  horizontal  piece  is  a  thrust ;  but  as  to 
which  way  the  arrow  indicating  the  stress  Q  in  the  oblique 
member  ("  web-member")  should  be  pointed  (a  matter  not  always 
to  be  decided  on  mere  inspection)  the  detail  of  the  analysis  would 
show,  provided  numerical  data  were  given  throughout.  Examples 
of  the  application  of  Hitter's  method  will  be  given  later. 

This  method  is  peculiarly  well  adapted  to  the  treatment  of 
pin-connected,  straight-linked,  trusses,  since  the  stress  in  a  straight 
link  (its  own  weight  being  neglected)  is  always  a  simple  tension 
or  compression  ;  in  other  words,  a  pull  or  thrust  directed  along 
the  axis  of  the  piece. 

The  three  stresses,  P,  Q,  and  T,  which  were  stated  to  be 
obtainable  from  the  free  body  in  Fig.  31,  in  the  three  straight 
links  concerned,  could  also  be  determined  from  the  consideration, 
as  a  free  body,  of  the  other  portion  of  the  truss,  viz.,  that  on  the 
right  of  the  cutting  surface  A  . .  B  of  Fig.  30 ;  in  factj  in  the 
present  case  with  greater  simplicity,  since  in 
this  new  free  body,  shown  in  Fig.  32,  the  sys- 
tem of  forces  in  equilibrium  is  much  simpler, 
though  there  is,  of  course,  the  same  number  of 
unknown  quantities,  P,  Q,  and  T\  which,  it  is 
to  be  carefully  noted,  are  the  equals  and  oppo- 
FIG.  32.  rites,  respectively,  of  the  P,  Q,  and  T  of  Fig.  31. 

31.  Remark. — From  the  foregoing  illustrations  (dealing  with 
compound  quiescent  structures  bearing  loads)  we  note  that  in  the 
various  free  bodies  (whose  conception  has  been  necessary  for 
introducing  the  different  unknown  forces  into  balanced  systems) 
the  pressures  or  pulls  of  any  two  parts  of  the  same  free  body 
against  each  other  are  omitted  from  the  system  ;  and  also  (see  last 
paragraph)  that  a  portion  of  a  two-force  piece,  situated  at  one 
end  thereof,  may  be  conceived  removed,  and  the  pull  or  thrust 
of  that  portion  against  the  remaining  portion  inserted  in  the 
system,  provided  it  is  a  straight  two-force  piece,  whenever  it  is 
desired  to  consider  separately  either  part  of  a  pin-connected  truss, 


PROBLEM   OF  THE  TWO   LINKS. 


33 


conceived  to  be  divided  into  two  parts  by  a  cutting  plane  or  sur- 
face (see  the  last  three  figures).  (The  stresses  in  bodies  that  are 
not  straight  links  will  be  considered  later  in  the  "Mechanics  of 
Materials,"  pp.  195-514,  M.  of  E.) 

32.  Problem  of  the  Two  Links  (Prob.  2  and  Fig.  36  of  p.  35, 
M.  of  E.). — Consider  link  AB  free  in  Fig.  33.  Since  this  is  not 
a  two-force  piece,  the  action-lines 
of  the  hinge-pressures  at  the  ex- 
tremities do  not  coincide,  nor  does 
either  follow  the  axis  of  the  piece. 
Hence  these  pressures  are  best 
represented  by  their  horizontal 
and  vertical  components,  as  shown, 
so  that  the  four  unknown  quanti- 
ties, X0 ,  Y0 ,  X1 ,  and  Y1 ,  are  to 
be  determined.  Similarly,  considering  the  other  link  as  free,  we 
have  Fig.  34,  in  which  it  is  to  be  noted  that  the  components  of 
the  pressure  at  the  upper  hinge  are  respectively  equal  and  oppo- 
site to  those  (X^  and  I7,)  of  Fig.  33  (action  and  reaction),  so  that 
there  are  only  six  unknown  u  force-amounts"  in  the  two  figures, 
instead  of  eight,  as  might  appear  at  first  sight  Hence  the  prob- 
lem is  determinate,  since  from  each  of  these  free  bodies  we  obtain 
three  independent  equations;  or  six  in  all,  as  follows:  Putting 
2X  =  0,  2  Y  =  0,  and  2(Pa)  =  0  (i.e.,  2  moments),  for  each 
body  in  turn,  taking  the  centre  of  moments  at  the  lower  hinge  in 
each  case,  we  have 

For      Fig.  33  j       ,     *«  ~~  ^^  Q '>   YI  +^°  ~  ^  ""^  =  °5 
(  and  JLJi  -\-  1  ,0  —  Cr1a1  —  6r2«2  =  0. 


FIG.  34. 


For 


Fig.  34  |          T^       ^%7  f  ' ^       ^l       G*  ~  °  ; 
(  atid  7})'  —  X,ti  +  G3a9  =  0. 


The  elimination,  by  which  to  find  separately  the  six  unknowns, 
JT0,  yo,  JT/,  r,,  2Ta,  and  I7,,  is  left  to  the  student.     (Treated 


graphically,   this   case  would   come   under  Class  B  of  p. 
M.  ofE.) 


458, 


34 


NOTES   AND   EXAMPLES   IN"   MECHANICS. 


33.  Problem  of  Rod  and  Cord  (Prob.  4  and  Fig.  41  of  p.  37 

of  M.  of  E.). — Consider  the  rod  free  by 
cutting  the  cord  and  removing  the  pin 
of  the  hinge  ;  that  is,  besides  the  forces 
6rj  and  £r2,  we  must  insert  the  unknown 
tension  P  along  the  axis  of  the  cord,  at  an 
angle  a  with  the  rod,  and  the  horizontal 
and  vertical  components,  X0  and  Y0,  of  the  hinge-pressure,  whose 
action-line  is  unknown,  and  thus  have  a  complete  svstem  of  forces 
in  equilibrium.  There  are  only  three  unknown  quantities,  P\ 
X, ,  and  Y0 . 

From  2  hor.  comps.=0,  we  have  X0— P'coso^O  ;  .     .     .     (1) 

«     2  vert.       "     =0,  u       "     Y0+P'  sin  a-  6^-  #,=0 ;  (2) 

2  (moms,  about  hinge)  ...  P'c—  6>,  —  6X=0.    ...     (3) 

Since  eq.  (3)  contains  only  one  unknown,  P',  we  have  at  once 

Pf=(Glal  +  G,fi3  +  c; 

and  knowing  P ',  we  obtain  X0  from  (1),  and  Y0  from  (2) ;  and 
finally  the  hinge-pressure,  R  =  V X*  +  Y0*9  making  an  angle 
whose  tan  =  Y0  -r-  X0  with  the  horizontal. 

34.  Problem  of  Simple  Roof-truss  (Frob.  5  and  Fig.  40  of  p. 
37,  M.  of  E.).— The   right-hand 

support  is  supposed  to  furnish  all 
the  horizontal  resistance.  Hence 
the  system  of  forces  acting  on  the 
whole  truss,  considered  free,  will 
be  as  shown  in  Fig.  36,  in  which 
there  are  three  unknown  reac- 
tions (or  pressures,  of  the  support- 
ing surfaces),  V0 ,  Vn ,  and  H.  II 
becomes  known  from 

2X=  0,     viz.,     2  TFsin  a  -  H  =  0.   . 
By  moments  about  point  (9,  we  have 

VJ-PJ-P^tf-  Wb  +  0  +  0  +  0  =  0, 
which  can  be  solved  for  Vn ,  while  from  2  moms,  about  B 
_  yj  +  Pl+  W .  I  cos  a  +  W(l  cos  a  -  1)  +  P,  .  ±1  =  0,  (3) 

from  which  F0  can  be  obtained. 


FIG.  36. 


(i) 

(2) 


SIMPLE   KOOF-TRUSS. 


35 


[NOTE. — Having  now  made  use  of  three  independent  equations, 
based  on  the  laws  of  equilibrium  of  forces,  and  by  their  aid  de- 
termined three  quantities  originally  unknown,  the  student  should 
Dot  imagine  that  by  putting  2  Y  =  0,  or  by  writing  another 
moment  summation  about  the  point  A  (for  example),  he  thereby 
secures  another  independent  equation,  from  this  same  free  body, 
capable  of  determining  a  fourth  unknown  quantity.  He  would 
find  that  such,  an  equation  could  be  established  by  mere  algebra, 
from  the  first  three  above,  without  further  reference  to  the  figure, 
and  hence  would  be  useless  as  regards  determining  any  other 
unknown  quantities.  See  top  of  p.  33,  M.  of  E.] 

All  loads  or  forces  being  considered  to  act  at  the  joints, 
and  no  piece  extending  beyond  a  joint,  we  note  that  this  roof- 
truss  is  composed  entirely  of  straight  two-force  pieces  (each  in 
simple  tension  or  compression  along  its  - 
axis),  so  that  portions  of  the  truss  may  be 
considered  free,  isolated  by  the  passing  ot' 
one  or  more  cutting  surfaces.  For  ex- 
ample, to  find  the  stress  in  piece  AO  and 
that  in  CO  consider  the  free  body  in  Fig. 
37,  where  S  and  T  are  the  stresses  required, 
figure  form  a  concurrent  system,  for  which 


FIG.  8? 

The  forces  in  this 


=  0  gives  8  cos  or+Tcos  /?+  TFsin  a=0,     .     .     .    (4) 
and  2  Y  =  0     "      S  sin  a+T  sin  /?+  F0— P—  TFcos«r=0. .  (5) 

Solve  these  for  S  and  T.  In  a  numerical  case  one  or  both  of  these 
will  come  out  negative,  indicating  compres- 
sion, not  tension  as  assumed  in  the  figure. 
To  find  the  stresses  in  A  C  and  AB,  the 
free  body  shown  in  Fig.  38  may  be  taken. 
Here  the  forces  form  a  non-concurrent  sys- 
tem. Taking  moments  about  B,  we  have 
Ta-  U.  iZ  +  0  +  0  =  0;  .  (6) 
from  which,  T  being  already  known,  U  can 
be  obtained.  For  R,  put  2Y  =  0,  and  it  will  be  the  only  un- 
known quantity  ;  or,  put  moments  about  C  =  0. 


-_______•>, 


FIG.  38. 


36  NOTES   AND    EXAMPLES   IN   MECHANICS. 

35.  Problem.  Rod  and  Tumbler  (Fig.  39).— The  tumbler  is 
smooth-edged  with  vertical  sides.  The  rod 
has  smooth  sides  and  weighs  G  Ibs.  C  is 
its  centre  of  gravity,  at  a  distance  a  from 
the  end  of  rod.  Given  the  distances  a  and 
d,  at  what  inclination  a  with  the  horizontal 
should  the  rod  be  placed,  in  contact  with 
tumbler  at  two  points  as  shown,  that  the 
P°8iti°n  of  the  rod  may  be  stable,  i.e.,  that 
FlG-  39-  the  rod  may  remain  in  equilibrium  ?  G, 

a,  and  d  are  known ;  H,  P,  and  a  unknown  ;  R  and  P  being 
the  pressures  of  the  tumbler  against  the  rod  at  the  two  points  of 
contact.     II  must  be  horizontal  (why  ?),  and  P  1  to  side  of  rod 
and  hence  at  angle  a  with  the  vertical. 
The  rod  being  the  free  body, 

from  2Z  =  0  we  have  H  —  P  sin  a  =  0; (1) 

u     2  T  =  0    "      "      —  G  +  P  cos  a  =  0  ; (2) 

"     2  (morns,  about  0)  we  have  Pd  sec  a  —  Ga  cos' a  =  0.  .  (3) 
Now  sec  a  —  1  -~  cos  a,  and,  from  (2),  G  =  P  cos  a.     Hence 
(3)  becomes 

/7  3  [d 

P Pa  cos3  a  =  0 ;    .-.  cos  a  =  \  /  -. 

cos  a  V   a 

a  being  now  known,  P  and  H  are  easily  found  from  (2)  and 
(1).  (A  brief  mode  of  finding  a  alone  is  based  on  the  fact  that 
the  three  action-lines  concerned  must  meet  in  a  common  point  m. 
If,  therefore,  a  figure  be  drawn  in  which  the  action-line  of  P  in- 
tersects that  of  II  in  a  point  m'  not  coincident  with  m,  that  of  II 
and  G,  we  have  only  to  form  a  trigonometrical  expression  for  the 
distance  ~mm' ,  involving  «,  d,  and  a,  write  it  =  0,  and  solve  for 
cos  cf  or  sec  «.) 

36.  Problem.  Pole  and  Tie  (Fig.  40).— Given  the  load  P,  the 
weight  G  of  the  pole,  and  all  the  distances  and  angles  marked  in 
the  figure,  it  is  required  to  find  the  tension  P'  induced  in  the 
chain,  whose  weight  is  neglected,  and  which  thus  serves  as  a 
straight  tie.  The  pole  is  hinged  at  0. 

The  pole    may  be  considered  free,  as  already  shown  in  the 


POLE  AND   TIE. 


37 


Pole  and 


figure,  by  inserting  the  pull  P'  exerted  on  it  at  ~k  in  a  known 

action-line,  Jc  .  .  Z,  by  the   chain,  and  the 

horizontal   and    vertical   components,   XQ 

and  Y0  ,  of  the  hinge-pressure  at  0.     The 

action-line    of  this  hinge-pressure    makes 

an  unknown    angle   with  the   horizontal, 

but   must    pass    through    the   centre    of 

the  hinge   at    0.      There  are   three   un- 


knowns in  the  system,  viz.,  X0  ,  Y 
Pf. 

From  2X  =  0  we  have 


and 


FIG.  40. 

Pf  cos  a  -  X0  =  0  ;  .     .     (1) 

«     2Y=0   "     "       Y.-P-G—  P'sin  a  =  0-,.    .    (2) 

«      2  (moms,  about  0)  =  0,Pa  +  Gb  —P'a'  =  0.  .     .     (3) 

The  value  of  P'  is  easily  found  from  (3),  then  that  of  X0  from 

(1),   and    that   of    Y9   from   (2).      Hence    the   amount  of  the 

hinge-pressure,  P0  =  VX*  +  Y*,  becomes  known,  and  the  tan- 

gent, =  Y0  -f-  XQ  ,  of  the  angle  between  its  action-line  (On)  and 

the  horizontal. 

Graphic  Solution.  The  action-line  of  the  resultant,  It, 
=  P  +  6r,  of  the  known  parallel  forces  P  and  G  may  easily  be 
determined  by  a  construction  like  that  of  Fig.  10,  M.  of  E.  ;  or 
by  applying  the  principle  of  the  foot-note,  p.  14,  M.  of  E.  Since 
this  action-line  intersects  that  of  the  force  P'  in  some  point 
n,  the  hinge-pressure  at  0  must  act  along  the  line  O  .  .  n  and 
must  be  the  equal  and  opposite  of  the  resultant  of  P'  and  R. 
37.  Problem,  Three  Cylinders  in  Box  (Fig.  41).  —  Three  solid 
homogeneous  circular  cylinders,  of  equal 
weight,  =  G,  and  of  the  same  dimen- 
sions, rest  in  a  box,  as  shown,  of  hori- 
zontal bottom  and  vertical  sides.  Th^ 
two  lower  cylinders  barely  touch  each 
other  ;  i.e.,  there  is  no  pressure  between 
them,  as  the  box  is  an  easy  fit.  The 
centres  of  the  three  cylinders  form  the 
vertices  of  an  equilateral  triangle.  It  is 
required  to  find  the  pressures  at  all  points  of  contact  (points,  in 


Three  Cylinders. 


NOTES   AND   EXAMPLES   IN   MECHANICS. 


FIG.  42. 


FIG.  43. 


this  end  mew  ;  really,  lines  of  contact)  between  the  cylinders  and 
the  box ;  also  between  the  cylinders  themselves.  All  surfaces 
smooth. 

Fig.  42  shows  the  upper  cylinder  as  a  free  body,  there  being 
three  forces  acting  on  it,  viz.,  (?,  the  action 
of  the  earth,  or  gravity,  and  the  two  press- 
ures, P'  and  P")  from  the  cylinders  be- 
neath. From  symmetry,  Pf  and  P"  must 
be  equal.  From  2  (vert,  comps.)  =  0, 

2P'  cos  30°  —  G  =  0  ; 
i.e.,  P'  (or  P")  =  G  +  V3. 
Taking  now  the  lower  right-hand  cylinder  free  in  Fig.  43,  we 
find  it  under  the  action  of  its  weight  G ;  of  the  pressure  P'  (the 
equal  and  opposite  of  the  P'  in  Fig.  42)  now  known ;  of  the  un- 
known horizontal  pressure  or  reaction  P'"  from  the  side  of  the 
box ;  and  of  the  vertical  pressure  JP0  from  the  bottom,  also 
unknown.  (Concurrent  system,  with  two  unknown  quantities.) 
There  is  no  pressure  at  A.  (See  above.)  From  2  (hor. 
comps.)  =  0, 

P'  sin  30°  -  P'"  =  0 ;  .*.  P'"  =  G(  V$  -r-  6). 
From  2  (vert,  comps.)  =  0,P0-G  -Pr  cos  30°=  0 ;  .-.  P0=  f  #. 
It  thus  appears  that  the  sum  of  the  two  pressures  on  the 
bottom  of  the  box  is  3£r,  i.e.,  is  equal  to  the  combined  weight 
of  the  three  cylinders ;  if  the  sides  of  the  box  were  not  vertical, 
however,  this  would  not  be  true,  necessarily. 

38.  Problem  of  the  Door. — Fig.  44  shows  an  ordinary  door  as 
a  free  body.  Support  is  provided  by  two  verti- 
cal hinge-pins,  of  smooth  surfaces,  whose  dis- 
tance apart  is  such  that  the  lower  one  alone 
receives  vertical  pressure  from  the  door  (i.e., 
furnishes  a  vertical  supporting  force,  V0 ,  at  its 
horizontal  upper  face).  The  upper  hinge-pin  pro- 
vides only  lateral  support,  as  seen  in  the  horizon- 
tal reaction  H  which  prevents  the  door  from  fall- 
ing away  to  the  right,  from  that  hinge.  Simi- 
larly, the  right-hand  vertical  edge  of  the  lower 


FIG.  44. 


DOOR — WEDGE   AND   BLOCK. 


39 


hinge-pin,  by  its  reaction  H0,  offers  lateral  support  at  that  point. 
Given  the  weight  G  of  the  door   (considered  as  concentrated  in 
its  centre  of  gravity)  and  the  distances  a  and  A,  it  is  required  to 
determine  the  three  pressures,  H,  H0  ,  and  V0  . 
From  2  (vert,  comps.)  =  0  we  have  V0  —  G  =  0  ;  i.e.,   V0  —  6r. 


From  2  (moms,  about  0), 


—  #a  =  0  ;  i.e.,  H=-j-G. 


From   2  (hor.  comps.)  =  0,      J7  —  H0  =  0  ;  i.e.,  #~0  =     #. 

A 

Since  V9  is  parallel  and  numerically  equal  to  6r,  and  IT  to 
7/0,  the  system  of  forces  acting  on  the  door  is  seen  to  consist  of 
two  couples  of  equal  moments  of  opposite  sign,  thus  balancing 
each  other.  The  smaller  the  distance  h  the  greater  the  value  of 
H  (and  of  its  equal,  If0\  if  G  and  a  remain  unchanged. 

(Unless  the  fitting  of  the  parts  is  very  accurate,  only  one 
hinge  of  a  door  receives  vertical  pressure  —  i.e.,  carries  the  weight, 
in  practical  language.)  If  more  than  one  receives  vertical  press- 
ure, the  share  carried  by  each  depends  on  the  accuracy  of  fitting 
and  on  the  slight  straining  or  change  of  form  of  the  parts  under 
the  forces  acting.) 

39.  Problem  of  the  Wedge  and  Block  (Fig.  45).—  The  shaded 
parts  represent  a  smooth  horizontal  table 
or  bed-plate  eh,  and  a  flat  and  smooth  ver- 
tical guide  md,  both  immovable.  W  is 
the  wedge  whose  angle  of  sharpness  is  a, 
and  B  the  block,  on  which  rests  a  weight 
(not  shown)  whose  pressure  on  B  is  verti- 
cal and  =  Q.  The  weights  of  wedge  and 
block  are  neglected.  There  is  supposed  to 
be  no  friction,  otherwise  t!,c  results  would 
be  quite  different  (see  §  9  of  Graph.  Stat. 
of  Mechanism,  in  this  book,  and  p.  171,  M.  of  E.).  Given  Q 
and  a,  what  force  P  must  be  applied  horizontally  at  the  head  of 
the  wedge  to  prevent  the  block  B  from  sinking  (or  to  raise  the 
block  with  constant  velocity  if  the  latter  has  an  upward  motion)? 

Supposing  the  required  force  P  to  be  in  action,  and  that  there 


FIG.  45. 


40 


NOTES  AND   EXAMPLES   IN   MECHANICS. 


FIG.  46. 


FIG.  47. 


is  no  friction,  the  mutual  pressure,  N,  between  block  and  wedge 
is  normal  to  the  surface  of  contact  and  hence 
makes  an  angle  a  with  the  vertical,  while  the 
pressures  on  surfaces  cd  and  eh,  viz.,  8  and  J?, 
are  horizontal  and  vertical,  respectively.  Hence, 
when  the  block  B  is  considered  free  (Fig.  46)  we 
hrve  equilibrium  between  the  three  forces,  Q,  N,  and  /&;  the 
two  last  are  unknown,  but  are  determined  thus : 

From  2  (hor.    comps.)  =  0,     /S  =  JV"  sin  a ; 
From  2  (vert,  comps.)  =  0,     N  cos  a  =  Q. 
Similarly,  we  have  the  wedge  free  in  Fig.  47   under  the 
action  of  jV,  now  known,  and  the  unknown 
and  the  required  P.     Hence 

R  =  N  cos  a,  (=  Q\  from  2  Y  =  0 ; 
and    P  =  N  sin  a,     =  Q  tan  oc,  from  2X  —  0. 

Evidently,  the  sharper  the  wedge  the  smaller 
the  force  P  necessary  for  a  given  Q. 

40.  Cantilever  Frame  (Fig.  48). — This  frame  consists  of  eleven 
straight  pieces  in  a  vertical  plane, 
pin-connected,  and  supposed  without 
weight.  Pieces  (7,  Z>,  E,  and  /  are 
horizontal,  G  and  H  vertical,  the 
others  oblique.  The  vertical  rod  at 
A  is  anchored,  as  shown,  but  no  stress 
is  induced  in  it  (nor  in  any  other 
piece)  unless  a  load  P  is  placed  at  w 
(no  other  joint  to  be  loaded).  Since, 
furthermore,  no  piece  connects  more 
than  two  joints,  each  piece  is  a  straight  two-force  piece,  subjected 
to  a  tensile  or  compressive  stress  along  its  axis,  and  any  one  may 
be  conceived  to  have  a  portion  removed,  when  desired,  in  the 
isolation  of  the  various  "free  bodies"  to  be  considered.  The 
necessity  of  the  rod  and  anchorage  is  evident  from  the  fact  that 
the  rectangle  m'nom  is  left  unbraced.  The  load  P  being  given, 
it  is  required  to  find  the  stress  in  each  piece  of  the  frame  and  in 
the  rod  at  A. 


FIG.  48. 


WEDGE — CANTILEVER  FRAME. 


41 


FIG.  49. 


The  free  body  in  Fig.  49  enables  us  to  find  the  stresses  E 
and  F.     Since  only  three  forces  are  concerned, 
meeting  at  a  point,  a  simple  procedure  is  to 
resolve  the  known  P  into  two  components,  E'  *-* — ^  "/ffi""*^ 
and  F')  along  the  action-lines  of  E  and  F,  as       »     /&\    / 
shown  by  dotted  lines.     Er  and  Ff  are  equal 
and  opposite  to  the  required  stresses  E  and  f\ 

respectively ;    i.e.,    E  =  P  cot  a  =  P  j  ,  and 

/^=jPcosec  or=  P  -=-  sin  a.  (The  summation  of  horizontal  and 
vertical  components  put  equal  to  zero  would  give  the  same  result.) 
^is  tension;  F,  compression. 

Next  take  the  free  body  in  Fig.  50,  involving  the  known  force 
P  and  the  unknown  stresses  7>,  7,  and  Z, 
assuming  for  them  the  characters  implied  by 
the  pointing  of  the  arrows.  Taking  moments 
about  centre  of  pin  at  0,  we  have 


FIG.  50. 


-Pa  =  Q;    whence  _Z>  =  +  P, 
and  is  tension. 


From  2  (vert,  comps.),  P  =  L  cos  ft,  or  L  =  +  ~ 


COS  fJ 

is  therefore  compression  as  assumed  ;  while,  from  ^(hor.  comps.) 
=  0,  I—  D  —  L  sin  ft  =  0  ;  i.e.,  /"=  +  P  (tan  /?  +  cot  a\  and 
is  compression. 

To  find  the  stresses  in  A,  B,  and  G  we  use  the  free  body  in 
Fig.  51.     By  moments  about  O, 


m' 


e 


.  • .  A  =  -f-  P  -j ,  and  is  tension. 


X      f 

/M 

^-^-.^..VH 


From  2  (hor.  comps.)  =  0, 

D  —  ^cos  y  —  0; 

whence  B  =  4-  -7-  • ,  and  is  compression,  as  assumed. 

'  •  fi  cos  y  ' 


FIG.  51. 


42  NOTES   AND   EXAMPLES   IN   MECHANICS. 

In  Fig.  4$  we  note  that  since  pieces  C  and  D  are  in  the  same 
straight  line  and  G  is  the  only  other  piece 
connecting  with  the  joint  m'  (there  being  also 
no  load  on  m'\  the  stress  in  the  piece  G  must 
be  zero  and  the  stress  in  C  must  equal  that  in 
D  ;  similarly,  the  stress  in  piece  II  is  zero  and 
E=  D  (as  already  found  above).  Hence  G  is 
marked  =  0  in  Fig.  52,  showing  a  free  body 
FIG.  52.  in  which  the  stresses  J  and  K  are  the  only 

unknown  forces,  C  being  =  D,  =  P  j ,  and  G  =  zero.     There- 
fore, by  elimination  between  the  equations 

J  cos  0  -f-  C  —  I  —  R  cos  d  —  0  (from  sum  of  hor.  comps.),  and 
J  sin  0  -\-l£  sin  #  —  A  =  0   (from  sum  of  vert,  comps.), 
we  obtain  the  stresses  J  and  K,  both  of  which  are  assumed  to  be 
compressions  in  this  figure. 

If  the  joints  m'  and  n  were  not  both  in  the  horizontal  line 
joining  w  and  v,  Fig.  48,  the  stresses  in  G  and  H  would  not  be 
zero,  as  they  are  in  this  instance. 

40a.  Real  and  Ideal  Forces.  "  Balanced  Forces." — The  student 
should  be  careful  to  distinguish  between  real  and  ideal  forces. 
If  a  body  A  receives  pressures  P  and  Q  from  bodies  B  and  C  re- 
spectively, the  resultant  of  P  and  Q  is  purely  id-eal,  being  merely 
conceived  to  take  the  place  of  P  and  Q,  if  any  useful  and  legiti- 
mate purpose  can  thereby  be  served.  Again,  the  X  and  Y 
components  of  an  actual,  or  real,  pressure  P  are  ideal  ^  serving  a 
mathematical  purpose  only,  when  we  suppose  P  to  be  removed 
and  its  components  inserted  in  its  stead. 

Such  customary  phrases  as  "balanced  forces,"  "forces  in 
equilibrium,"  etc.,  are  unfortunately  worded,  as  they  seem  to 
imply  that  forces  act  on  forcos,  which  is  an  absurdity.  In  reality, 
bodies  act  on  bodies,  force  being  the  mere  name  given  to  the 
action  (if  it  is  a  push,  pull,  etc.) ;  so  that,  instead  of  stating  that 
"  the  forces  are  balanced,"  we  should  more  logically  say  :  "The 
rigid  body  is  in  equilibrium,  or  balanced,  under  the  actions  of 
certain  other  bodies."  (See  correspondence  in  the  London 
Engineer  of  June,  July,  and  Aug.,  1891.) 


CHAPTEK  III. 
MOTION  OF  A  MATERIAL  POINT. 

41.  Velocity  and  Acceleration.  —  Any  confusion  between  the 
ideas  of  velocity  and  acceleration  is  fatal  to  a  clear  understanding 
of  the  subject  of  motion.  Just  as  velocity  may  be  defined  as  the 
rate  at  which  distance  is  gained,  so  acceleration  may  be  defined 
as  the  rate  at  which  velocity  is  gained,  thus  :  If  at  a  certain 
instant  a  material  point  has  a  velocity  of  10  feet  per  second,  we 
mean  that  it  has  such  a  speed  of  motion  that  it  would  pass  over 
a  distance  of  10  feet  ^that  rate  of  speed  remained  constant  dur- 
ing the  whole  of  the  next  second  of  time.  Similarly  if,  at  the 
same  instant,  the  acceleration  of  the  material  point  is  said  to  be 
20  feet  per  "  square  second  "  (or  "  20  feet  per  second,  per  second  "), 
we  mean  that  if  the  velocity  were  to  continue  to  change,  during 
the  whole  of  the  next  full  second,  at  the  same  rate  at  which  it  is 
changing  at  the  instant  mentioned,  the  velocity  at  the  end  of  that 
second  would  be  greater  by  20  units  of  velocity  than  at  the  in- 
stant mentioned,  i.e.,  the  velocity  would  be  30  feet  per  second. 
According  to  this  definition,  then,  if  the  velocity  of  the  material 
point  remains  unchanged,  the  acceleration  is  zero  at  every  point 
of  the  motion. 

At  each  point,  then,  in  the  path  of  a  material  point  moving 
along  a  right  line,  we  have  to  do  with 

Sj  its  distance  from  some  convenient  origin  in  that  line  ; 

v9  its  velocity,  or  rate  at  which  that  distance  is  changing,  =  -.-  ; 

dt 

p,  its  acceleration,  or  rate  at  which  the  velocity  is  changing, 
dv 


There  is  no  need  of  defining  still  another  quantity  as  the  rate 
at  which  the  acceleration  is  changing,  for  the  reason  that  the 

43 


44  NOTES   AND   EXAMPLES   IN   MECHANICS. 

force  which  at  any  instant  occasions  the  peculiarity  of  the  motion 
of  the  material  point  is  determined  from  the  acceleration,  viz., 
from  the  relation  force  =.  mass  X  accel. 

In  the  uniformly  accelerated  motion  of  a  free  fall  in  vacuo, 
the  following  are  the  values  of  the  above  quantities  at  beginning 
of  the  motion,  and  at  the  end  of  each  full  second  thereafter,  thus  : 
s,  =  Distance,     v,  —  Yeloc.     p,  —  Acceleration  =  g. 
At  beginning,    ...      0.0  ft.      0.0  ft.  p.  sec.    32.2  ft.  p.  (sec.)2 ; 
At  end  of  1st  full  sec.,    16.1ft.    32.2     "      "       32.2     " 
At       "      2d     "     "       64.4ft.    64.4     "      "      32.2     "        « 
At       "      3d  '  "     "     144.9  ft.    96.6     «      "      32.2     "        " 

A  recent  English  writer  is  so  desirous  that  acceleration  shall 
not  be  confused  with  velocity  that  he  calls  the  unit  of  velocity  a 
"  Speed  "  and  the  unit  of  acceleration  a  "  Hurry"  For  example, 
using  the  foot  and  second  as  units,  he  would  say  that  at  the  end 
of  the  second  full  second  of  a  free  fall  in  vac.uo  the  velocity  is 
64.4  "speeds"  while  the  acceleration  at  the  same  instant  is  32.2 
"hurries."  These  words  are  quite  suggestive  and  should  be 
borne  in  mind. 

To  say  that  at  a  certain  instant  the  acceleration  is  zero  does 
not  imply  that  the  body  is  not  moving,  but  simply  that  its  veloc- 
ity, however  small  or  great,  is  not  changing;  and  again,  the 
statement  that  the  velocity  is  zero  at  a  certain  instant  does  not 
imply  that  the  acceleration  is  zero  also,  but  only  that  the  velocity, 
as  its  value  changes,  is  then  passing  through  the  value  zero,  while 
the  rate  at  which  it  is  changing  (i.e.,  the  acceleration)  is  not  de- 
termined until  some  further  statement  is  made. 

42.  Momentum. — This  word  is  used  quite  largely  in  some 
works  on  Mechanics,  but  may  be  considered  a  superfluity,  liable 
to  give  rise  to  confusion  of  ideas ;  though  sometimes  useful,  it 
need  rarely  be  used.  By  definition,  it  is  a  name  given  to  the 
product  of  the  mass  of  a  material  point  by  its  velocity  at  any 
instant,  i.e.,  Mv.  Of  course,  the  mass  of  the  body  is  a  constant 
quantity,  while  its  velocity  may  be  continually  changing;  hence 
the  momentum  is  always  proportional  to  the  velocity.  In 
accordance  with  this  definition,  the  value  of  a  force  which  is 
accelerating  the  velocity  of  a  material  point  in  its  path  is  some- 


MOMENTUM — COKD   AND    WEIGHT.  45 

times  stated  to  be  equal  to  the  rate  at  which  the  momentum  is 
changing  ;  by  which  is  simply  meant  the  following: 

From  eq.  (IV),  p.  53,  M.  of  E.,  we  have  P  =  Mp,  =  (mass  X 

dv 

accel.),  butj?  =  -j-  ;  hence  we  may  write 
cut 

^dv      Mdv      d\_Mv\  _  ("change  of  momen- 
dt  ~  '    dt  dt       ~  L    turn  in  time  dt 

i.e.,  P  =  the  rate  at  which  the  momentum  is  changing.  There- 
fore the  above  statement  as  to  the  value  of  the  accelerating  force 
is  nothing  more  than  what  we  already  have  in  the  form  of 
P  =  Mp. 

43.  Cord  and  Weights. — There  are  few  mistakes  more  common 
than  rushing  to  the  conclusion  that  the  tension  in  a  vertical  cord 
to  which  a  weight  is  attached  is  equal  to  that  weight.  This  may 
be  true  (and  is  true  if  the  weight  is  at  rest  and  has  no  other  sup- 
port), but  should  not  be  assumed  without  thought. 

For  example,  Fig.  53,  having  two  weights  attached  to  the  same 
cord,  if  the  point  A  of  the  cord  were  fastened,  A 

the  tension  in  B  would  be  =  G  ;  and  that  in 
(7,  =  G',  if  G  and  G'  are  the  respective 
weights  of  the  two  bodies.  But  if,  the  cord 
bein^  continuous  and  not  fastened,  with  no 
friction  at  the  pulley-axles,  an  accelerated 
motion  begins  (assume  G'  >  G),  the  tension 
on  each  side  depends  on  that  acceleration,  FIG.  53. 

as  well  as  on  the  weights  of  the  bodies.  To  find  this  accel- 
eration, which  is  common  to  both  sides,  neglecting  the  masses  and 
weights  of  the  pulleys  and  cord  (by  which  we  mean  that  the 
tension  S  in  the  cord  at  A  may  be  taken  equal  to  that  at  B  and 
C}  let  us  consider  the  body  G'  free.  We  note  that  this  body  has 
a  downward  accelerated  motion,  and  that  the  forces  acting  on  it 
are  6r',  directed  vertically  downward,  and  S,  the  tension  in 
the  cord,  pointing  vertically  upward,  i.e.,  acting  as  a  resistance; 

G' 
hence,  calling  the  acceleration  p,  we  have  G'  —  /S  =  — p.     As 

t7 

for  the  other  weight,  it  is  rising  with  an  upward  acceleration  =  p. 


46  NOTES   AND   EXAMPLES   IN   MECHANICS. 

under  an  upward  force  8  and  a  downward  resistance  G,  whence 
S  —  G  =  —p.     From  these  two  relations  we  obtain  by  elimina- 

y 


tion 

G'  -  G 


and     *= 


The  value  of  p  might  have  been  obtained  directly  by  consid- 
ering that  the  acceleration  of  the  motion  is  just  the  same  as  if  the 

G+G' 

whole  mass  --  :  -  were  moving  in  the  same  right  line  under 

i/ 

the  action  of  a  single  accelerating  force  Gr  —  G. 

44.  Example.  Lifting  a  Weight.—  A  rigid  mass  weighing  100 
Ibs.  is  to  be  lifted  vertically  through  a  distance  of  80  ft.  in  4  sec- 
onds of  time,  and  with  uniformly  increasing  velocity,  from  a 
condition  of  rest  (i.e.,  veloc.  —  0).  "What  tension  must  be  main- 
tained in  a  vertical  cord  attached  to  it,  to  bring  about  this  result? 

The  average  velocity  is  20  ft.  per  second,  and  since  the  initial 
velocity  is  zero  and  the  rate  of  increase  of  velocity  (i.e.,  the 
acceleration,  p)  is  to  be  constant  (uniformly  accelerated  motion), 
the  final  velocity  will  be  double  the  average,  viz.,  40  ft.  per  sec. 
(see  also  eq.  (4),  p.  54,  M.  of  E.).  If,  then,  40  velocity-units  are 
gained  in  4  seconds,  the  acceleration  is  40  -=-  4  =  10  ft.  per  second 
per  second  (10  "  hurries"),  and  an  upward  accelerating  force  of 

Mp  =  2^2  X  10  =  31  Ibs.   must  be  provided.     But  the  only 

forces  actually  present,  acting  on  this  mass,  are  the  action  of  the 
earth,  viz.,  100  Ibs.  pointing  vertically  downward,  and  the  ten- 
sion, S,  in  the  cord,  directed  vertically  upward  ;  and  their  (ideal) 
resultant,  which  is  S  —  100  Ibs.  (since  they  have  a  common  action- 
line),  is  required  to  be  31  Ibs.  and  to  act  upward  ;  whence  we 
have  the  required  cord-tension,  =  S,  =  131  Ibs. 

To  lift  the  100-lb.  weight,  then,  under  the  conditions  imposed, 
requires  a  cord-tension  of  131  Ibs.  [If  the  cord  be  allowed  to 
become  slack  at  the  end  of  the  80-ft.  distance,  the  body  does  not 
immediately  come  to  rest,  as  it  then  has  an  upward  velocity  of 
40  ft.  per  second.  Its  further  progress  is  an  "  upward  throw" 


HAKMONIC   MOTION.  47 

(p.  52,  M.  of  E.)  with  40  ft.  per  second  as  an  initial  velocity,  the 
only  force  acting  at  this  stage  being  the  downward  attraction  of 
100  Ibs.,  which  will  gradually  reduce  the  velocity  to  zero.] 

If  a  smaller  height  of  lift  had  been  assigned  with  the  given 
time,  or  the  same  height  with  a  longer  time,  the  difference  be- 
tween the  requisite  tension  S  and  the  gravity  force  (or  weight)  of 
100  Ibs.  would  have  been  smaller ;  and  vice  versa. 

It  is  convenient  to  note  that  in  using  the  foot-pound-second 
system  of  units  for  force,  space,  and  time,  the  mass  of  a  body  is 
obtained  by  multiplying  its  weight  in  pounds  by  0.0310  (which 
is  the  reciprocal  of  32.2) ;  thus,  in  the  preceding  example  the 
mass  of  the  100-lb.  weight  is  =  100  X  .0310  —  3.10  mass-units. 

45.  Harmonic  Motion. — From  eq.  (3),  p.  59,  M.  of  E.,  remem- 
bering that  G  and  a  are  constants,  we  note  that  s,  the  "  displace- 
ment," or  distance  of  the  body  from  the  origin  (middle  of  the 
oscillation),  is  proportional  to  the  sine  of  an  arc  or  angle,  and  that 
this  arc  is  proportional  to  the  time,  t,  elapsed 
since  leaving  the  origin  ;  whence  arise  the 
following  graphical  relations :  In  Fig.  54,  let 
O  be  the  origin  or  middle  of  the  oscillation,  c, 
and  the  horizontal  line  CD  the  path  of  the 
body  ;  OD,  =  /»,  being  the  extreme  displace- 
ment, i.e.,  the  "  semi-amplitude,"  of  the  mo- 
tion. At  any  instant  of  time  the  body  is  at 
some  point  m  between  C  and  D,  i.e.,  Om  =  the  variable  s  (dis- 
placement). With  0  as  centre,  and  OD,  =  r,  as  radius,  describe 
a  circle  and  erect  a  vertical  ordinate  at  m  at  whose  intersection  n 
with  the  curve  draw  a  radial  line  On,  making  some  angle  6  with 
the  vertical,  and  the  abscissa  nk.  Now  0m,  or  s,  =  r  sin  8  ;  and 
the  length  of  curve  En  is  proportional  to  the  angle  0.  Hence 
the  linear  arc  En  is  proportional  to  the  time  occupied  ~by  the 
body  in  describing  the  distance  s  =  Om  ;  that  is,  if  n  be  re- 
garded as  a  moving  point,  confined  to  the  circumference  of  the 
circle,  and  always  in  the  vertical  through  m,  its  motion  being 
thus  controlled  by  the  harmonic  motion  of  w,  the  velocity  of  n  in 
its  circular  path  is  constant  (equal  linear  arcs  in  equal  times). 
The  constant  velocity  of  n  must  be  equal  to  the  initial  velocity 


48  NOTES   AND   EXAMPLES   IN   MECHANICS. 

of  ra,  i.e.,  c,  as  the  latter  leaves  the  origin,  0\  for  at  0  the  two 
points  are  both  moving  horizontally. 

Conversely,  therefore,  if  a  point  n  move  in  the  circumference 
of  a  circle  with  a  constant  velocity  c  (continuously  in  one  direc- 
tion), the  foot  of  its  ordinate,  i.e.,  the  point  m,  moves  with  har- 
monic motion  along  the  horizontal  diameter ;  this  is  the  proof 
called  for  in  the  line  below  Fig.  64  of  M.  of  E.  The  0  of  Fig. 
54  is  t  Va  of  the  formulae  on  p.  59,  M.  of  E. 

Since  the  acceleration  (as  also  accelerating  force)  in  harmonic 
motion  is  proportional  to  the  displacement,  the  length  of  the  line 
0m,  or  kn,  represents  the  acceleration  at  any  instant,  while  the 
length,  mn,  of  the  ordinate  is  proportional  to  the  velocity  of  the 
body  or  material  point,  m  (since  by  eq.  (4)  of  p.  59,  M.  of  E., 
that  Velocity  varies  as  the  cosine  of  0). 

Of  course,  in  obtaining  values,  from  the  drawing,  of  these 
variables  -y,  p,  and  £,  for  different  positions  of  the  body  m,  due 
regard  must  be  paid  to  the  scales  on  which  the  lengths  marked 
in  the  figure  represent  these  variables,  none  of  which  is  a  linear 
quantity. 

46.  Numerical  Example  of  Harmonic  Motion  (using  the  foot, 
pound,  and  second). — With  the  apparatus  and  notation  of  p.  58, 
M.  of  E.,  suppose  that  by  previous  experiment  with  the  elastic 
cords  we  find  that  a  tension  Tl  of  4  Ibs.  is  required  in  either  of 
them  to  maintain  an  elongation  of  3  in.,  i.e.,  £  of  a  foot ;  then 
for  any  elongation  s  (or  "  displacement"  of  the  body  during  its 
motion)  the  tension  (and  retarding  force)  is  T  =  T,s  -=-*,  =  16,9. 
Let  the  small  block  weigh  128.8  Ibs.,  then  its  mass  is  128.8  X  .0310 
=  4  =  M ;  and  hence  the  constant  quantity  called  a  is 
a  =  (Tv  ~  Ms^)  =  16  —  M ;  i.e.,  a  =  4. 

Let  an  initial  velocity  of  c  =  4  ft.  per  sec.,  from  left  to  right, 
be  given  to  the  block  at  0 ;  required  the  extreme  distance  attained 
by  the  block  from  0  (i.e.,  the  semi-amplitude,"  =  /•),  and  the  time 
occupied  in  describing  the  semi-amplitude  (i.e.,  required  the  time 
of  a  quarter-period). 

From  p.  59,  M.  of  E.,  r  =  c  ~  \f~cu;  .-.  r  =  4  ~-  V~±  =  2  ft. ; 
while  for  a  quarter-period,  or  half-oscillation,  we  have  the  time 
\n  -r-  Va  =  %7t  =  0.7854  sec. 


BALLISTIC   PENDULUM.  49 

Hence  in  the  diagram  of  Fig.  54  we  make  OD  ==  r  =  2  f t. 
Then,  since  the  linear  arc  ED  represents  0.7854  seconds,  the  time 
represented  by  En  for  any  position  m  of  the  block  will  be  on  a 
scale  of  £  sec.  to  each  foot  of  En  (since  0.7854  -f-  (\7t  x  2  ft.)  =  J). 
Since  OE,  or  2  ft.,  represents  the  velocity  of  m  on  leaving  O, 
i.e.,  4  ft.  per  second,  mn  represents  the  velocity  at  m  on  a  scale 
of  2  ft.  per  second  to  each  foot  of  mn.  Similarly,  since  the 
acceleration  OD  of  m  at  D  (from  jp  —  —  as)  is  —  (4  X  2)  =  —  8 
ft.  per  (sec.)2,  that  at  m  will  be  kn  on  a  scale  of  4  to  each  foot  of  kn. 

It  must  be  noted  that  after  m  reaches  D  and  begins  to  return 
toward  the  left,  the  point  n  in  the  curve  will  have  passed  below 
the  horizontal  through  D ;  that  is,  the  time  is  now  greater  than 
ED,  and  the  velocity  has  changed  sign.  When  n  passes  under- 
neath 0  the  acceleration  and  the  displacement  change  sign  ;  and 
so  on  indefinitely.  (Tension  in  cord  at  D  —  ?) 

47.  The  Ballistic  Pendulum. — This  old-fashioned  apparatus  for 
determining  the  velocity  of  a  cannon-ball  consisted  of  a  heavy 
body  B,  of  mass  Mz ,  suspended,  by  a  rod 
or  rods,  from  a  horizontal  hinge  on  a  fixed 
support.  A  cavity  in  its  side  is  partly  filled 
with  clay,  so  as  to  make  the  impact  inelastic. 
This  body  being  initially  at  rest  (see  Fig.  55), 
the  ball  is  shot  horizontally  into  the  cavity,  in 
which  it  adheres.  As  a  result  of  the  impact 
the  centre  of  gravity  of  the  combined  masses 
receives  a  velocity  C,  with  which  it  begins  its 
ascent  along  the  circular  arc  J3m<,  finally  reaching  a  vertical 
height  II  above  its  initial  position  before  coming  to  rest  (for  an 
instant).  H  being  measured,  or  computed  from  the  observed 
angle  ft,  we  have  O=  +  VZyH  (see  foot  of  p.  80,  M.  of  E.). 
The  known  mass  of  the  ball  being  M1  and  its  unknown  velocity 
just  before  impact  -|-  cl ,  eq.  (4)  of  p.  65,  M.  of  E.,  enables  us  to 
write 


(In  this  theory  the   suspended  mass  has  been  treated  as  a 
material  point,  and  the  result  is  only  approximate.     The  longer 


50  NOTES   AND   EXAMPLES   IN   MECHANICS. 

and  lighter  the  suspension-rod  and  the  smaller  the  dimensions  of 
the  suspended  body  the  more  accurate  the  results.  The  strictly 
correct  theory  is  rather  complicated.) 

48.  Apparatus  for  the  Determination  of  the  u  Coefficient  of 
Restitution."  —  It  will  be  noted,  by  an  examination  of  eqs.  (2),  (3), 
(5),  and  (6),  on  pp.  64  and  65,  M.  of  E.,  that  the  coefficient  qf 
restitution,  <?,  may  be  defined  as  the  ratio  of  the  loss  of  momentum 
of  the  first  body  in  the  second  period  of  the  impact  (period  of 
expansion  or  restitution)  to  the  momentum  lost  by  it  in  the  first 
period  (compression)  ;  and  similarly  for  the  second  body,  except 
that  for  it  there  is  a  gain  of  momentum  in  both  periods,  instead 
of  a  loss  ;  hence 

F2  -  C) 


e  = 


and  also    e  = 


Fig.  56  shows  the  apparatus.     The  two  balls,  of  same  sub- 
stance, are   suspended   by  cords  so 
that  they  can  vibrate  in  the  same 
vertical   plane.     When    hanging  at 
rest  they  barely  touch,  without  press- 
\6/    ure>  w^h  cen*res  at  tne  same  level. 
Being  allowed  to  swing  simultane- 
ously from  rest  at  a  and  J  respec- 
tively, their  impact  takes  place  at  O 
FIG.  56.  (or  very  nearty  so),  their  velocities 

just  before  impact  being  cl  =  +  VZg^  and  <?2  =  —  V2gh9 ,  re- 
spectively (see  foot  of  p.  80,  M.  of  E.),  where  \  and  A2  are  the 
vertical  heights  fallen  through  (each  ball  having  underneath  it  a 
graduated  arc,  so  that  each  of  these  heights  can  be  computed 
from  the  observed  angle  and  known  length  of  cord).  Supposing 
each  ball  to  rebound  on  its  own  side  of  0  and  the  heights  reached, 
Ht  and  H^  to  be  noted  or  computed,  their  velocities  at  0  im- 
mediately after  impact  must  have  been  T7,  =  —  VZgH^  and 
ya  =  -|-  V%g //,  respectively,  A  and  B  being  the  points  reached 
at  the  ends  of  the  rebounds.  Knowing,  then,  k,  and  A2,  Ml  and 
M9 ,  II,  and  H, ,  we  compute  e  from  either  eq.  (7)  or  (8)  of  p.  65, 
M.ofE. 


SPRING   BETWEEN   TWO   BALLS.  51 

Of  course  e  is  always  less  than  unity,  and  due  regard  must  be 
paid  to  the  signs  of  the  velocities  in  making  the  computations. 

49.  Action  of  a  Spring  between  Two  Balls  (Fig.  57). — In  the 
second  period  of  a  direct  central  impact 
(period  of  expansion,  or  restitution)  the  dis- 
tance between  the  centres  of  the  two  masses 
is  increasing,  the  front  body  being  subjected 
to  a  forward  pressure  precisely  equal  to  that 
which  at  the  same  instant  is  retarding  the 
hinder  body.  Hence  the  gain  in  momentum 
of  the  front  body  during  this  period  is  Fl«-  r>?- 

equal  to  the  loss  in  momentum  of  the  other  body ;  i.e.  (see  eqs. 
(5)  and  (6),  p.  65  of  M.  of  E.),  M,(  F2  -  <7)  =  M}(C  -  V,). 

This  is  precisely  what  takes  place  when  a  spring  in  a  state  of 
compression  (the  ends  tied  together  by  a  cord)  is  placed  endwise 
between  the  two  suspended  balls  of  the  apparatus  of  Fig.  57,  and 
the  cord  is  then  severed  without  violence  (burning  is  best).  The 
spring  in  regaining  its  natural  length  exerts  equal  horizontal 
pressures  in  opposite  directions  at  any  instant  against  the  two 
masses.  This  pressure  is  variable,  but  at  any  instant  has  the  same 
value  for  one  mass  as  for  the  other.  That  is,  this  phenomenon 
may  be  looked  upon  as  a  case  of  impact  where  the  first  period  is 
lacking,  the  period  of  restitution  occupying  the  whole  time  of 
impact.  In  this  case  Ml  and  J/3  are  at  rest  just  before  impact, 
that  is,  £7—0;  while  their  velocities  after  the  expansion  of  the 
spring  are  —  V^gH^  and  +  1/2<///2  respectively  (if  Hl  and  H^  are 
the  respective  vertical  heights  reached  by  the  balls  as  the  result 
of  the  action  of  the  spring).  Hence,  from  the  above  equation, 
J/2(+  VZgH>  —  0)  =  Mt(0  —  [—  VtyHA) ;  i.e.,  the  velocities 
immediately  after  the  action  of  the  spring  are  inversely  propor- 
tional to  the  masses. 

The  same  method  in  another  form  consists  in  saying  that  from 
the  principle  of  the  conservation  of  momentum  the  total  momen- 
tum before  impact,  viz.,  0,  here,  is  equal  to  that  after  impact, 
which  is  M,  V,  +  M,  F2,  or  M,(  -  VtyH?)  +  J/2(  +  V%glQ. 
Equating  the  latter  expression  to  the  zero  momentum  first  men- 


52  NOTES   AND   EXAMPLES  IN  MECHANICS. 

tioned,  we  have  Ml  VZgll,  —  Jf2  V2gJJ^9  as  before.     This  result 
is  verified  by  the  apparatus. 

50.  The  Cannon  as  Pendulum. — As  the  converse  of  the  Ballis- 
tic Pendulum,  the  cannon  itself  may  be  suspended  in  a  horizontal 
position  from  a  pivot  6',  Fig.  58.  With  the  ballistic  pendulum 
the  impact  was  inelastic,  i.e.,  it  consisted  of  a 
period  of  compression  only,  that  of  restitution 
being  lacking.  Here,  however,  the  expansion 
of  the  hot  gases  generated  when  the  powder 
is  ignited  acts  like  the  spring  in  the  preceding 
|DM2  case,  causing  a  forward  pressure  at  any  instant 
^*7*  against  the  ball  and  an  equal  and  simultaneous 
backward  pressure  against  the  cannon ;  i.e., 
the  impact  is  one  having  no  period  of  compression  but  only  one 
of  restitution.  The  velocity  F2  of  the  ball  leaving  the  muzzle 
may  therefore  be  inferred  from  a  knowledge  of  the  two  masses 
concerned  and  the  height  of  recoil  77,  (A  being  the  point  where 
the  centre  of  gravity  of  the  gun  comes  to  rest  for  an  instant). 
Putting  the  total  momentum  before  impact,  which  is  zero  (cannon 
at  rest,  in  lowest  position),  equal  to  that  immediately  after,  the 
cannon  having  then  just  begun  its  motion  from  D  toward  A  with 
a  velocity  at  D  of  Vl  =  —  V^gl^ ,  we  have 


or      ,  = 

As  before,  we  here  treat  the  masses  as  material  points. 

51.  Simple  Circular  Pendulum  of  Small  Amplitude  (Fig.  59). — 
B  is  a  material  point  suspended  from  a  fixed 
point  C  by  an  imponderable  and  inextensible 
cord.  Being  allowed  to  sink  from  initial  rest  at 
the  point  A  (cord  taut),  it  follows  the  circular 
arc  ABO  with  increasing  velocity.  At  any 
point  B  between  A  and  0  its  velocity  (which 
of  course  is  tangent  to  the  curve)  is 

v  =  V%g  X  vertical  height  I)F~ 
(see  foot  of  p.  80,  M.  of  E.),  and  the  forces  acting  on  it  are  its 
weight  G,  vertically  downward,  and  the  tension  S,  in  the  cord. 
This  tension  increases  as  the  body  approaches  O,  since  2  (norm. 


SIMPLE   0IKCT7X,Afi  PENDULUM.  53 

comps.)  must  —  Mtf  -=-  rad.  of  curv.,  [eq.  (5),  p.  76,  M.  of  E.],  i.e., 
#_£cos0:=:  —  .^;    whence   S=  ^fcos  0+^-1.     .     (1) 

y    "  jfw-j 

At  (9  the  tension  is  greatest  and  =  £r    1  +  -j-   .     If  the 

starting  point  A  is  taken  high  enough,  the  cord  may  be  broken 
before  0  is  reached.  To  find  the  tangential  acceleration  pt9  or 
rate  at  which  the  velocity  is  increasing,  we  note  that  from  eq.  (5), 
p.  76,  M.  of  E.,  2  (tang,  comps.)  must  =  Mpt,  whence 

Mpt  =  G  sin  0  -f-  S  X  0  ;     or     pt  =  g  sin  0.     .     (1) 


This  can  be  written  pt  =  |-  •  Z  <sm  0  =  ^-  •  ^^.     ....     (2) 

But  if  the  angle  AGO  is  very  small  (not  over  5°,  say),  BF 
may  be  put  —  linear  arc  OB\  and  as  0  is  the  middle  of  the 
oscillation,  and  BO  is  the  distance  or  displacement  of  the  body 
from  0  at  any  instant,  (2)  may  be  stated  in  the  form  :  the  accelera- 
tion is  proportional  to  the  displacement',  so  that  the  motion  is 
(very  nearly)  harmonic  (see  p.  58,  M.  of  E.).  To  find  the  time  of 
passage  from  A  to  0  on  this  basis  (half-oscillation),  note  that  on 


p.  59,  etc.,  M.  of  E.,  the  time  of  a  half-oscillation  is  -$ 


, 
Va 


where  a  is  the  quantity  which  multiplied  by  the  displacement  s 
gives  the  acceleration.  Hence  from  eq.  (2)  above,  the  "a"  of  the 
present  harmonic  motion  is  g  ~-  I.  Hence  the  time  of  a  whole 
oscillation  from  A  to  the  corresponding  extreme  point  on  the 
other  side  of  0  is 


-:t 

The  duration  of  the  oscillation  is  independent  of  the  amplitude 
(with  above  limitations).  (For  a  large  amplitude  see  foot  of  p. 
81,  M.  of  E.) 

With  an  extensible  cord,  elastic  or  inelastic,  the  results  would 


be  quite  different.  In  such  a  case  the  relation  v  =  %g  .  DJf  for 
the  velocity  at  any  point  B  would  not  be  true,  since  the  ten- 
sion S  is  not  perpendicular  to  the  velocity  when  the  cord  is 
elongating. 


CHAPTER  IY. 

NUMERICAL  EXAMPLES  IN  STATICS  OF  A  RIGID  BODY  AND 
DYNAMICS  OF  A  MATERIAL  POINT. 

52.  Example  1.    Anti-resultant  of  Two  Forces  (Fig.  60).— Two 

forces  in  a  horizontal  plane,  P  and  Pf, 
R  are  respectively  3  tons  (short  tons,  of  2000 
Ibs.  each),  and  6000  Ibs.  P  is  directed  North 
30°  East;  and  P',  South  85°  East.  The 
angle  between  them  is  therefore  65°.  Re- 
quired the  amount  and  position  of  R ',  their 
anti-resultant  (i.e.,  the  force  that  will  bal- 
ance them).  R'  must  be  equal  and  opposite 
to  the  (ideal)  resultant,  7?,  of  the  two  forces.  Adopting  the  short 
ton  as  a  unit  for  forces,  we  note  that  P'  and  P  are  equal,  each 
being  3  tons.  Hence  the  parallelogram  of  forces  formed  on  them 
as  sides  is  a  rhombus,  and  R  bisects  the  angle  between  ;  whence 
'we  may  write 

R=ZP  cos  32£°  =  2  x  3  x  0.8434  =  5.06  tons. 

Therefore  an  anti-resultant  of  5.06  tons,  directed  South  62J°  West 
must  be  provided,  if  the  two  given  forces  are  to  be  balanced  by 
a  single  force. 

53.  Example  2.     Resultant  Couple  of  a  Number  of  Couples.— 
Six  couples  acting  on  a  rigid  body  and  in  the  same  plane  (a  ver- 
tical plane)  have  the  following  moments,  respectively,  as  seen 
from  the  west  side  of  the  plane : 

+  30  ft.-lbs. ;         + 196  f t.-oz. ;        +  0.48  inch-tons ; 
—  0.08  ft-tons  ;     —  160  ft.-lbs.;  and  —  1300  inch-lbs. 

Required  the  moment  of  their  resultant  couple. 

54 


ETC.  55 

Changing  the  form  of  these  various  moments  to  what  they 
would  have  been  if  all  the  various  forces  had  been  expressed  in 
Ibs.  and  the  arms  in  feet  (i.e.,  expressing  them  all  in  ft.-lbs.),  and 
adding  those  of  the  same  sign,  we  have 

+  (30  +  12.25  +  80)  ft.-lbs.         =  +  122.25  ft.-lbs. ; 
and      -  (160  +  160  +  108.33)  ft.-lbs.  =  —  428.33      u 

—  306.08      " 

Since  the  algebraic  sum  of  the  moments  is  —  306.08  ft.-lbs. 
(the  couples  being  in  the  same  plane  (see  §  34,  M.  of  E.),  this  is 
the  moment  of  their  (ideal)  resultant  couple.  To  hold  the  given 
couples  in  equilibrium,  therefore,  a  single  couple  (their  anti- 
resultant)  having  a  moment  of  +  306.08  ft.-lbs.  must  be  applied  * 
to  the  body  to  preserve  equilibrium.  That  is,  if  a  seventh  couple 
in  which  (for  example)  each  force  is  153.04  Ibs.,  with  an  arm  of 
2  ft.,  and  appearing  counter-clockwise  seen  from  the  west,  be 
applied  to  the  body,  the  seven  couples  will  balance. 

If  the  given  couples  were  not  all  in  the  same  plane,  or  in 
parallel  planes,  we  should  have  to  make  use  of  the  results  of  §  33, 
M.  of  E. 

54.  Example  3.  Centre  of  Gravity  of  Trapezoid  with  Semicircle 
Cut  Out  (homogeneous  thin  plate  of  uniform  thickness — Fig.  61). — 
The  diameter  of  the  semicircle  lies  on 
the  lower  base  OK  of  the  trapezoid,  H  Y  /" 
is  the  centre  of  gravity  of  the  semi- 
circle (so  that  II D  =  4r  -4-  3?r),  E  that 
of  the  complete  trapezoid,  and  E'  that 
of  the  actual  plate.  See  figure  for 
given  numerical  dimensions  and  for 
notation  of  co-ordinates  of  centres  of  w[<— 6-(i-->p2T)" 
gravity  of  actual  plate  and  complete  FlG-  61- 

trapezoid,  denoting  those  of  the  semicircle  by  a?2  and  y2 .  Re- 
quired a?!  and  yl ,  using  the  inch  as  linear  unit.  The  area  of  the 
trapezoid  is  F  =  \  X  6  X  (10  +  16)  =  78  sq.  in. ;  that  of  the 
semicircle  is  F^  —  i7r(2)2  =  6.28  sq.  in. ;  so  that  the  area  of  the 
actual  plate  is  their  difference,  F^ ,  =  71.71  sq.  in.  Also,  for  the 
point  //  we  have  ya  =  4  X  2  ~  BTT  =  0.85  in. ;  and  OJ)  =  x, 

*  III  the  same,  or  iu  a  parallel,  plane. 


«--• 


56  NOTES   AND   EXAMPLES   IN   MECHANICS. 

=  8  in.  ;  while  (see  p.  23,  M.  of  E.)  y  =  |  .||:  1~  =  2.77  in.; 

and  x  =  OD  —  $  X  2.77  =  7.54  in. 

(C  bisects  the  base  MN\  conceive  perpendiculars  let  fall 
from  C  and  E  upon  OK,  and  use  the  similar  triangles  so  formed 
in  obtaining  the  value  of  a?.) 

From  eq.  (3),  p.  19,  M.  of  E.,  since  the  trapezoid  is  made  up 
of  the  actual  plate  and  the  semicircle,  we  have 
x  =  (F&  +    X)  ±  (F>  +  jg  ^ 

-       Fy-F,y, 

and      =  -          - 


By  substitution,  therefore, 

OY  =  [78  X  7.54  -  6.28  X  8]  +  71.71       =  7.5    in.  ;  and 
yv  =  [78  X  2.77  -  6.28  X  0.85]  -~  71.71  =  2.94  " 
55.  Example  4.     Stability  of  Two  Cylinders.  —  Fig.  62  gives  an 
TWO  cylinders,      end  view  of  two  smooth  and  ho- 

Equal  lengths  -,  *•      i  /. 

mogeneous  circular  cylinders  of 
equal  length  (—  I'  in  feet)  but  of 
radii  1  in.  and  3  in.,  respectively. 
A  weighs  800  Ibs.  per  cub.  ft., 
E  only  100  Ibs.  per  cub.  ft.     A 
and  B  being  placed,  as  shown,  on 
two  smooth  planes  at  45°  with  the 
^•2  -------  horizontal,  it  is  required  to  find 

whether  this  is  a  stable  position, 
or  if  A  will  crowd  B  out  of  place.  Call  their  total  weights  G,  and  £ra 
for  the  present.  Since  CD  is  one  half  of  OC,  the  angle  COD  is 
30°.  First,  the  position  being  supposed  stable,  to  find  the  press- 
ure at  point  2,  we  take  A  as  a  free  body.  The  forces  acting  on  it 
are  three,  shown  at  0  in  Fig.  63.  Pl  is  the  pressure  (or  reaction) 
of  the  inclined  plane  against  A  at  point  1,  P2  is  the  pressure 
from  the  other  cylinder,  and  Gl  the  weight  of  this  cylinder,  A. 
These  are  all  directed  through  0  (smooth  surfaces),  the  angles 
being  as  shown.  For  equilibrium  P2  must  be  equal  ai.d  opposite 
to  OK,  the  (ideal)  resultant  of  Gl  and  Pl  .  Jn  triangle  OPf, 
OK\  G,  ::  sin  45°  :  sin  60°  ;  .-.  OK=  G[  ^2  -~  1/3]  =  P%. 


Rad.  of  Cyl 

A  is  1  inch.        $ 


NUMEEICAL   EXAMPLES — STATICS.  57 

In  Fig.  63  we  have  acting  through  C  the  four  forces  acting 
on  the  large  cylinder  B  (on  supposition  —R 

of  equilibrium).     Gt  is  its  weight,  Pz  is 
the   pressure  of   the  other  cylinder,  P3 
the  pressure  at  point  3  of  the  inclined 
plane  on  the  right,  P4  of  that  on  the         ^      *P*       yG2 
left.  FIG.  63. 

If  now  the  resultant  of  P2  and  G9  were  found  to  pass  above 
the  point  3,  instability  would  be  proved;  since  to  occasion 
pressure  at  point  4  that  resultant  should  evidently  pass  below  3. 
Or,  which  amounts  to  the  same  thing,  assuming  equilibrium  at 
C  and  with  P4  as  drawn,  if  we  compute  the  value  of  P4  by  put- 
ting 2  (compons.  -|  to  P3)  =  0,  and  a  negative  result  is  obtained, 
instability  is  proved  ;  and  vice  versa.  Hence  we  write 

C1          C1 

P4  =  G,  cos  45°  -  P2  cos  60° ;  whence  P4  =  —^  -   — L.  .  (1) 

v  2       \  6 


Now  G,  =  |>(TV)3  X  800Z']  Ibs. ;  and  G,  =  [n(f^  X  100Z']  Ibs., 
on  substituting  which  in  (1)  we  obtain 

P4  =  [^rf'(  + 309.8)]  Ibs.; 

which  is  positive,  and  thus  verifies  the  supposition  of  equilib- 
rium. 

56.  Example  5.  Toggle-joint  (Fig.  64). — The  two  straight 
links,  of  equal  length,  are  pivoted  to  the  two 
— y —  blocks,  as  indicated.  A  horizontal  pull  of  80  Ibs., 
\ftt  making  equal  angles  with  the  two  links,  is  exerted 
on  the  horizontal  pin  of  the  joint  B.  What  press- 
ures are  thereby  induced  (for  given  position  of 
parts)  on  the  surfaces  E  and  F  (horizontal  and 
vertical)?  (Those  on  E'  and  F'  will  be  the 
same,  respectively.)  Each  link  is  evidently  a 
straight  two-force  piece  and  hence  under  a  com- 
pressive  stress  along  its  axis ;  call  this  stress  P '. 
The  free  body  in  Fig.  65  enables  us  to  find  P' 
from  ^  (hor.  comps.)  =  0;  i.e.,  2Pf  cos  a  =  P  ; 


58 


NOTES   AND   EXAMPLES   IN   MECHANICS. 


Fig.  66  shows  the  upper  block  free,  from  which  by 

PE  =  P'   sin    a,    and   from 
comps.  PF  =.  P'  cos  a. 


n  a= 


—  0, 
hor. 


=4SO  Ibs.  ; 


and 


If,  in  Fig.  65,  P  were  resolved 
into  components  along  the  axes  of 

f  &  \    the  two  links,  each  such  component 

/p  j       '      "B  would  be  the  equal  and  opposite  of 

FIG.  65.  FIG.  66.         the  corresponding  P ' .     Evidently, 

if  a  approached  a  right  angle.  P ',  and  also  PE,  would  increase 
without  limit. 

57.  Example  6.     Simple  Crane. — The  simple  crane  in  Fig.  67 
carries  a  load  of  4  tons  at  (7, 
12  ft.  from  the  axis  of  the  I       ^^c 

vertical  shaft,  while  its  own 
weight  is  1  ton,  the  centre 
of  gravity  being  3  ft.  from 
the  shaft.  The  socket  at  B 
is  shallow  so  that  lateral 
support  is  provided  at  A. 
Required  the  pressure  at  A 
and  the  horizontal  and  ver-  Fl°- 67-  FlG- 68- 

tical  pressures  from  side  and  bottom  of  socket  B.  The  crane 
being  considered  free  in  Fig.  68,  and  the  reacting  pressures  being 
put  in,  as  shown,  we  have,  from  2  (moms,  about  B)  =  0  (foot 
and  ton), 

AXX  8  —  4x  12—  1X3=0:   whence  Ax  =  6.37  tons. 


From  5"  (vert,  compons.),  .Z^  —  5  =  0  ;   or 
2  (hor.  compons.),  Bx  —  Ax  =  0  ;   or 


By  =  5  tons. 
Bx  =  6.37  tons. 


58.  Example  7  Door  and  Long  Hinge-rod. — The  door  weighs 
200  Ibs.,  and  is  supported  in  a  vertical  plane  in  the  manner 
indicated  in  Fig.  69.  The  continuous  vertical  rod  ABC  is  con- 
sidered without  weight,  and  from  the  nature  of  the  mode 
of  support  of  the  door  receives  horizontal  pressures  at  each 


NUMERICAL   EXAMPLES — STATICS. 


59 


of  the  points  A,B,  6",  and  O.  Kequired  the  values  (J.,  B,  <?', 
and  C)  of  these  forces  ;  also  the  vertical  pressure  C"  between 
the  projecting  shoulders  C  and  C'. 


FIG.  69.  FIG.  70. 

Fig.  70  shows  the  door  free,  and,  by  moments  about  Of,  using 
the  pound  and  foot, 

Ga  =  Ab ;  or  A  =  (200  X  1.5)  -s-  5  =  60  Ibs. 

2  (vert,  comps.)  gives 
C"  =  G  =  200  Ibs. ;  while  from 

2  (hor.  compons.),  —  ^L  -j-  C'  =  0,  whence 
C"  =  A  =  60  Ibs. 

The  rod  as  a  free  body  is  shown  in  Fig.  71  and  enables 
us  to  find  the  pressures  B  and  (7,  now  that  A  and  C'  are 
known.  By  moments  about  0,  we  have 

.Z?  X  4  —  60x5=0;  whence  j£  =  75  Ibs. 

In  putting  2  (hor.  compons.)  =  0,  since  C'  =.  A 
from  Fig.  70,  the  summation  reduces  to  C—B  =  0, 
i.e.,  C  =75  Ibs. 

The  hinge-rod,  therefore,  is  seen  to  be  under 
the  action  of  two  couples  of  equal  and  opposite 
moments  ;  one  consisting  of  A  and  C',  the  other 
of  B  and  C,  while  the  sill  C,  Fig.  69,  receives  a  vertical  pressure 
equal  to  the  weight  of  the  door. 

59.  Example  8.     Shear-legs   (Fig.    72).— The    weight    G  of 


FIG.  71. 


60 


NOTES   AND   EXAMPLES   IN   MECHANICS. 


G=  2000  76s. 
,=  600  Ibs. 


U 


FIG.  72. 


2000  Ibs.  is  supported  by  the  two  straight  links  in  a  vertical  plane 
as  shown,  with  given  dimensions  and  weights.  Required 
the  pressures  produced  on  the  hinge-pins  at  A  and  B.  If  the 
links  had  no  weight  they  would  be  straight  two-force  pieces,  CB 

being  subject  to  a  compression 
and  AC  to  tension  ;  and  the 
hinge-pin  pressures  at  A  and  B 
would  be  equal  to  these  forces, 
the  action-li  n  es  of  the  latter  bei  ng 
the  axes  of  the  pieces,  respective- 
ly. In  that  case  a  simple  solution 
would  consist  in  resolving  the 
force  of  2000  Ibs.  by  a  parallelo- 
gram into  two  components,  one  along  CB,  the  other  along  AC 
prolonged,  and  these  components  would  be  found  to  be  3367  and 
1813  Ibs.,  respectively;  the  former  being  the  compression  in  CB 
and  the  latter  the  tension  in  CA.  But  the  weights  of  the  links 
are  considerable  and  are  to  be  considered ;  hence  the  links  are 
not  two-force  pieces  and  must  not  be  conceived  to  be  cut  in  form- 
ing any  free  body.  The  hinge-pressures  at  A  and  B  are  replaced 
by  their  horizontal  and  vertical  components,  as  shown,  Ax  and 
Ay ,  Bx  and  By ;  these  four  are  the  unknown  quantities  re- 
quired. 

The  figure  shows  all  the  forces  acting  on  a  free  body  consisting 
of  the  two  links  and  the  2000-lb.  weight.  By  taking  moments 
about  A  we  exclude  three  of  the  unknown  quantities  and 
obtain 

+  2000  X  30  +  600  X  26  +  900  X  16  -  By  X  20  =  0  ; 

or  By  =  4500  Ibs. ; 
while  by  moments  about  B, 

2000  X  10+  600  X  6  -  900  X  4  -  Av  X  20  =  0  ; 
or  Ay  =  1000  Ibs. 

Now  conceive  the  link  CA  alone  to  constitute  a  free  body, 
the  forces  acting  being  Ax ,  Ay ,  Gt ,  and  the  pressure  of  the 


NUMERICAL   EXAMPLES — STATICS. 


61 


hinge-pin   at  C  against  this  link.     The   action-line  of  this  last 
force  is  not  known,  but  the  moment-sum  about  C  excludes  the 
force  and  gives 
+AX  X  20  —  1000  x  30  —  900  X  14  =  0 ;  whence  Ax  =  2130  Ibs. 

Since  in  the  first  free  body  Bx  must  =  Ax  (from  2  hor. 
coinps.  —  0),  we  have  also  Bx  =  2130  Ibs.  and  can  now  compute 
the  actual  oblique  hinge-pressures  A  and  B,  at  A  and  B  re- 
spectively. From 

A  =  VA;  +  A,',    and    B  =  VB?  +  Bv\ 
we  have  finally 

A  =  2354  Ibs.,          and    B  =  4978  Ibs. 

From  tan  "  \AV  -H  Ax)  we  find  that  A  makes  a  smaller  angle  with 
the  horizontal  than  the  link  CA  ;  and  similarly,  that  of  B  is 
greater  than  that  of  link  CB. 

60.  Example  9.  Roof  Truss  with  Loads  and  Wind  Pressures 
(Fig.  73). — Here  the  half-weight  of  each  piece  is  supposed  to  be 


2~tons 


FIG.  73. 

carried  directly  on  the  pin  of  the  corresponding  joint,  so  that 
each  link  or  member  will  be  considered  as  a  straight  two-force  piece 
and  hence  in  simple  compression  or  tension  along  its  axis.  In 
obtaining  free  bodies,  therefore,  any  piece  or  pieces  may  be  con- 
ceived to  be  cut  and  the  stress  inserted.  The  load  given  at  each 
joint  includes  the  half-weights  of  all  the  pieces  meeting  there. 
For  all  distances  and  angles  needed  see  figure.  The  wind  is 
supposed  to  blow  from  the  left,  its  pressure  (4  tons)  on  the  left 


62  NOTES   AND   EXAMPLES   IN   MECHANICS. 

slope  of  the  roof  being  normal  to  the  same,  half  borne  at  each 
joint,  b  and  c.  Resistance  to  horizontal  displacement  is  supposed 
to  be  provided  at  the  right  support,  alone ;  the  other  extremity 
of  the  truss  being  on  rollers,  so  that  the  reaction  there  is  vertical ; 
hence  at  the  right  we  have  two  reactions  to  deal  with,  horizon- 
tal and  vertical ;  i.e.,  Hn  and  Vn. 

Required  the  three  supporting  forces,  Hn  ,  Vn ,  and  F0 ;  and 
also  the  stresses  A,  (7,  D,  and  E  (and  their  character),  in  the 
pieces,  A,  C,  D,  and  E. 

Fig.  73  shows  the  whole  truss  as  a  free  body.     By  moments 
about  joint  &,  adopting  the  foot  and  ton  as  units,  we  have 
+  Fn  X  38  -  2  x  38  -  3  X  12  -  3  X  26  -4x19  -2  X  17.3  =  0; 
whence  Vn  =  7.91  tons.     From  2  (vert,  comps.)  =  0, 

K  +  Vn  —  2  —  3  —  3  —  4  —  2  —  2  sin  a  —  2  sin  A •  =  0 ; 
and  hence  F0  =  8.88  tons ;  while  2  (hor.  comps.)  =  0  gives 
+  2  cos  a  +  2  cos  a  —  Hn  —  0  ;     or     IIn  =  2.86  tons. 
Next,  considering  free  the  portion  of  the  truss  on  the  left  of 
a  plane  cutting  pieces  A,  D,  and  C,  we 
have  Fig.  74,  in  which,  for  the  present, 
we  assume  A  to  be  tension  and  C  and  D 
compression.     2  (moms.)  about   point  c 
(intersection  of  C  and  D)  gives 
^X  10.2+2X12+2X17.3-8.88X12=0; 
jv~8.88tons.  whence  A  =  +4.7  tons,  and  this  being 

Fl°- 74-  positive,  the  assumption  of  tension  is  con- 

firmed. 2  (moms.)  about  #,  similarly,  gives 
0X9  +  3X7  —  2X1.7 +  2X19 +  2X  15.6  —  8.88  X  19  =  0  ; 
or,  0  =  +  9.1  tons,  and  is  therefore  compression.  Although  the 
same  free  body  would  serve,  let  us  determine  stress  D  from 
another  free  body,  that  in  Fig.  75,  showing  the  remainder  of  the 
truss.  Assume  D  compression. 

From  2  (vert,  comps.)  =  0  we  have 

+  7.91  -3  —  2  —  4-  Asm/3-Dcosy  =  0; 
or,  D  =  —  2.45  tons.     The  negative  sign  showing  the  assumption 
of  compression  to  be  incorrect,  D  is  2.45  tons  tension. 


NUMEEICAL   EXAMPLES. 


63 


Again,   from   the   free   body   in    Fig.  76,   taking   moments 


Vn=  7.91  tons,     "n 
FIG.  75. 


FIG.  76. 


about  n,  having  assumed  Eto  be  tension,  we  have 

+  #x  17  +  3X12 -9.1  X  12.5  =  0; 
i.e.,  E=  +  4.56  tons,  and  is  tension. 

From  the  same  free  body  the  stress  in  G  is  easily  found. 

61.  Remark. — The  foregoing  examples  of  this  chapter  have 
all  involved  the  equilibrium  of  rigid  bodies,  each  under  a  system 
of  forces  in  a  plane.     Those   remaining  to  be  given,  however, 
deal  with  moving  material  points,  or  bodies  small  enough  to  be 
so  considered  (dynamics  of  a  material  point) ;  and  the  concurrent 
forces  in  each  case  acting  on  the  body  when  considered  free,  do 
not  form  a  balanced  system  (unless  the  motion  is  rectilinear  and 
of  constant  velocity),  so  that  ^2X  and  2  Y  are  not  =  0  neces- 
sarily.    For  example,  if  the  path  is  a  straight  line  which  is  taken 
as  the  axis  X,  then  ^X '=  mass  X  accel. ;  while  2  (comps.  -\  to 
the  path)  —  0,  as  if  the  forces  were  balanced.     But  if  the  path  is 
a  curve,  then  at  each  point  2  (comps.  along  the  tangent)  =  mass 
X  tan.  ace.  and  2  (comps.  along  the  direction  of  the  normal) 
=  mass  X  square  of  veloc.  4-  radius  of  curvature.     In  numeri- 
cal substitution  the  student  is  very  apt  to  forget  that  if  q,  the 
acceleration  of  gravity,  be  denoted  by  the    number  32.2,  times 
must  be  expressed  in  seconds  and  distances  iufeet.     (The  expres- 
sion for  the  mass  of  a  body  always  involves  the  quantity  g.) 

62.  Example  10.     Train  Resistance. — If  the  frictional  resist- 
ance of  a  certain  200-ton  railroad  train  be  assumed  to  be  equivalent 
to  a  backward  force  of  12  Ibs.  per  ton  applied  directly  to  the  car- 
frames  at  any  ordinary  speed,  in  what  distance  on  a  level  track 
will  the  train  be  stopped  if  moving  initially  at  a  velocity  of  40 


64 


NOTES  AND   EXAMPLES   IN  MECHANICS. 


miles  per  hour?  (There  are  no  brakes  on,  nor  any  locomotive; 
the  resistance  being  due  to  the  rubbing  of  the  journals  in  their 
boxes  and  the  unevenuess  and  compressibility  of  the  track  and 
wheel-treads  (rolling  resistance)  ). 

Ditto  :  if  the  train  is  on  an  up-grade  of  26.4  ft.  to  the  mile  ? 

Taking  the  axis  +  JTin  the  direction  of  motion,  we  note  that 
the  accelerating  force  is  —  2400  Ibs.  ;  i.e.,  that  the  sum  of  the 
comps.  along  the  path  is  —  2400  Ibs.  The  mass,  in  the  ft.-lb.-sec. 
system  of  units,  is  =  G  -+-  g  —  400,000  -r-  32.2.  Now  2X  =  Mp 
—  mass  X  ace.,  and  therefore  the  ace.  =p  —  2zX-±M  —  —  0.193. 
The  accelerating  force  being  constant,  the  acceleration  is  constant 
and  hence  the  motion  is  uniformly  accelerated  (retarded  here), 
and  the  eq.  (3)  of  p.  54,  M.  of  E.,  is  applicable,  viz,  distance  =  s 
=  (v*  —  c2)  -T-  %p.  The  initial  velocity  =  c  —  40  miles  per  hour, 
=  58.6  ft.  per  sec.,  while  v  is  to  be  zero.  Hence 


On  the  up-grade,  the  path,  or  axis  X,  is  inclined  upward  at  an 
angle  a  with  the  horizontal  (whose  tangent  is  26.4  -h  5280  =  -^ 
and  is  practically  =  sin  a)  and,  besides  the  —  2400  Ibs.,  the  X 
component  of  6r,  viz.,  —  G  sin  a  =  -g-j-g-  of  400,000  Ibs.  = 
—  2000  Ibs.,  acts  to  retard  the  motion. 
/.  2X=  —  4400  Ibs.  and  p  =  —  0.354  ft.  per  sec.  per  sec. 

.-.  s  =  [O2  -  (58.6)2]  -*-  [2  X  (-  0.354)]  =  4858  ft. 
63.  Example   11.     Inclined  Plane,   Two   Weights,    and  Cord 
(Fig.  77).  —  The  cord  connecting  the  two  weights  is  very  light  and 
inextcnsible,  and  friction   and   mass  of  the 
pulley  are   neglected.     (By  neglecting  the 
mass  or  inertia  of  the  pulley  we  mean  that, 
notwithstanding  the  fact  that  its  rotary  mo- 
tion is  accelerated  by  the  cord,  the  tension  in 
B   the  cord  is  the  same  at  any  instant  where  it 
FIG.  77.  leaves,  as  where  it  winds  upon,  the  pulley- 

rim.)  The  two  blocks  being  at  rest  in  the  position  shown  (cord 
taut,  with  a  temporary  support  under  JB\  the  support  is  sud- 
denly removed  ;  required  the  distance  sa  through  which  B  then 


NUMERICAL  EXAMPLES—  DYNAMICS  OF  MATERIAL  POINT.   65 

sinks  in  the  first  two  seconds  of  time  (=  &,),  and  also  the  tension 
in  the  cord  during  the  motion,  if  the  body  A  encounters  a  fric- 
tional  resistance,  on  the  inclined  plane,  always  equal  to  -fa  of  the 
normal  pressure  on  the  plane. 

Consider  A  free  at  any  instant  of  the  motion  (Fig.  Y8).  Call 
its  acceleration  pt  .  The  forces  acting  on 
it  are  its  weight  G,  S  the  tension  in  the 
cord,  the  friction  F^  and  the  normal 
pressure  N^  from  the  inclined  plane. 
(That  is,  the  resultant  of  Nl  and  Fis  the 
resultant  action  of  the  plane  on  body  A,  FlG-  78- 
which  resultant  action  is  evidently  not  normal  to  the  plane,  which 
it  would  not  be  *  unless  the  bodies  were  smooth.)  Although  the 
path  of  A  is  straight,  consider  it  as  a  particular  case  of  a  curve  ; 
then  (see  §  61  above) 

2  (tang,  comps.)      =  S  —  F  —  G  cos  a  =  —pt  ;....(!) 

t/ 

Mv*  Mv* 

2  (normal  comps.)  =  -  —  ;     or    ^  —  G-  sin  a  —  -  '—  =  0.    (2) 

From  (2)  we  find  the  value  of  N^  ;  and  hence 

F=*fNl=+,GA*a  .......    (3) 

At  this  same  instant  (which  is  any  instant  of  the  motion),  con- 
sider B  free  in  Fig.  79.  There  are  only  two  forces  acting  on  it  : 
its  weight  Gl  ,  and  the  upward  tension  Sl  in  this  part  of  the  cord. 
B  is  sinking  with  some  acceleration^?. 

From  2  (downward  comps.)  =  mass  X  ace.,  we  have 


But  (from  above  remark  on  mass  of  pulley,  etc.)  we  know  that 
$!  =  /$;  and  since  the  cord  is  taut  and  does  not  stretch,  p  must 
=  pt  .  (Let  the  student  devise  a  strict  proof  of  this.) 

Hence  by  elimination  between  the  four  equations  we  obtain 

&        F-  Gcosa 
___  *=*•  =  g+g.        ^  _ 

*  That  is,  not  necessarily. 


66  NOTES  AND   EXAMPLES  IN   MECHANICS. 

which  is  constant  and  hence  the  motion  is  uniformly  accelerated 
and  the  equations  of  §  56,  M.  of  E.,  hold  good,  among  which  is 
s  =  \ptf  when  the  initial  velocity  is  zero. 

Passing  to  numbers,  in  the  ft.-lb.-sec.  system,  we  have 

12-10x0.707(0.3  +  1)  /Q_ 
p  = 1Q   ,  12  -  •  (32.2)  =  4.11  ft.  per.  sec.  per.  sec., 

and  hence  from  (4), 

tension  =  S  =  12  [l  -  |^|]  =  10.47  Ibs. ; 

while  s,  =  tytf  =  i  X  4.11  X  (2)2  =  8.22  feet  =  distance  de- 
scribed in  the  first  two  seconds  (the  velocity  at  end  of  which 
—  vz  =  pi*  =  8.22  ft.  per  sec.). 

It  is  seen  that  the  weight  B  sinks  with  about  one  eighth  the 
acceleration  of  a  free  fall. 

64.  Example  12.  Free  Fall. — A  stone,  allowed  to.  descend 
freely  and  vertically,  from  rest,  occupies  -^  of  a  second  (17 
watch-ticks,  say)  in  falling  through  the  height  of  a  cliff ;  required 

this  height.     From  eq.  (2),  p.  51 ,  M.  of  E. ,  we  have  s  —  ct  +  ^— . 

2 

In  the  present  case  c  —  0,  and  hence  the  height  required  =  0  -f- 
32. 2  X  ix(Y-)2  =  290.8ft. 

On  account  of  atmospheric  resistance,  which  is  neglected  in 
the  theory  of  p.  51,  M.  of  E.,  and  is  variable  (being  nearly  pro- 
portional to  the  square  of  the  velocity  for  the  same  body),  the 
actual  height  is  smaller,  the  discrepancy  depending  on  the  shape, 
the  specific  gravity,  and  absolute  size  of  the  falling  body.  If  the 
stone  is  round,  and  about  one  inch  in  diameter,  the  average 
resistance  in  the  above  case  might  be  as  much  as  one-quarter  of 
its  weight,  so  that  we  might  write  %g  instead  of  g  for  a  rough 
approximation.  (See  p.  822,  M.  of  E.)  If  it  were  two  inches  in 
diameter  (same  substance),  its  weight  would  be  increased  eight- 
fold, and  the  average  resistance  about  quadrupled,  and  thus  the 
latter  might  be  about  one  eighth  of  the  weight. 

"With  smaller  heights  of  fall  the  resistance  is  much  smaller, 
not  only  absolutely  but  proportionally,  on  account  of  the  smaller 
average  velocity. 


NUMERICAL  EXAMPLES  —  DYNAMICS  OF  MATERIAL  POINT.  67 

65.  Example   13.    Block  on   Circular  Guide  (Fig.  80).—  The 
AD  smooth-  curved  guide  ABD  is  smooth  and  fixed, 
DE  rough.    of  ^he  form  of  the  quadrant  of  a  circle 
with  a  horizontal  tangent  at  D.     The 
plane  DE  is   rough.      The   block    G 
weighs  20  Ihs.  and  is  to  slide  from  rest 
FIG.  so.  at  A  down  the  circular  guide.     How 

far  (i.e.,  distance  sl  =  ?)  will  it  slide  on  the  rough  plane  DE 
before  being  brought  to  rest,  if  the  latter  offers  a  frictional  re- 
sistance of  20  oz.  (i.e.,  li  Ib.)  ?  The  radius  of  the  curve  in  which 
the  centre  of  G  moves  is  48  inches. 

The  velocity  vl  of  G  on  its  arrival  at  D  is  the  same  as  if  it 
fell  freely  through  the  corresponding  vertical  height  CD  =  48 
in.  ==  4  ft.  (the  time  of  descent,  however,  is  quite  different)  ;  for 
the  guide  is  both  fixed  and  smooth  (see  p.  83  and  also  foot  of 
p.  80,  M.  of  E.).  Adopt  the  foot,  Ib.,  and  second. 

.\  v,  =  V2  X  32.2  X  4  =  16.05  ft.  per  sec.  For  the  motion 
on  DE,  vl  is  the  initial  velocity,  and  the  motion  is  uniformly  re- 
tarded (i.e.,  the  acceleration  is  constant  and  negative)  if  we  take 
the  direction  from  D  toward  .Z^as  positive;  since  the  only  force- 
component  along  the  path  is  —  1.25  Ibs.,  the  gravity-force  of  20 
Ibs.  being  i  to  the  path.  The  acceleration  is^>  =  force  -f-  mass, 
the  mass  being  =  20  -f-  32.2  =  20  x  0.0310  =  0.620  ;  .-.  p  = 
(—  1.25)  -=-  .62  —  —  2.012  ft.  per  sec.  per  sec.,  and  from  eq.  (3) 
of  p.  54,  M.  of  E.  (in  which,  for  present  purposes,  we  put  s  =  81  , 
v  —  0,  c  =  vl  ,  and  p  as  above),  we  have 

*,  =  [—  (16.05)2]  -f.  [2(-  2.012)]  =  64  ft. 

If  we  inquire  the  pressure  P  between  the  block  and  curved 
guide  just  before  reaching  Z>,  we  note  that  that  pressure  must 
not  only  support  the  weight  of  the  body,  but  must  also  provide  a 

proper  deviating  force  -—  —  to  retain  it  on  the  curve  ; 


whence       P  =  20  Ib,  +  i-       Ib,  =  60  Ibs. 

(See  p.  83,  M.  of  E.)     The  pressure  on  DE  is  only  20  Ibs. 


68 


NOTES   AND   EXAMPLES  IN  MECHANICS. 


Again,  suppose  the  block  to  start  from  rest  at  .Z?,  the  angle 
BCD  being  45°  ;  find  vl  and  sl  .  (The  acceleration  on  DE  is  the 
same  as  before.) 


vl  =  V%  X  32.2  x  4(1  —  cos  45°)  =    8.68  ft.  per  sec. 
8l  =  (O2  -  <)  -T-  [2(-  .  2.012)]     =  18.75  ft. 

66.  Example  14.  Harmonic  Motion  of  a  Piston  (Fig.  81).  — 
On  account  of  the  great  mass,  and  large 
radius,  of  the  rim  of  a  fly-wheel  on  the 
same  shaft,  the  rotation  of  the  crank  is 
practically  uniform;  at  least  during  any 
one  turn,  i.e.,  the  crank-pin  is  considered 
to  move  with  a  uniform  velocity  in  a  cir- 
cle. From  the  design  of  the  piston  and 
slot  this  body  oscillates  with  harmonic  motion  in  a  horizontal 
path  (see  foot  of  p.  59,  M.  of  E.)  ;  the  left  to  right  stroke  alone 
is  to  be  considered.  The  pressure  called  P  is  the  total  effective 
steam-pressure,  i.e.,  the  difference  between  the  total  pressure  of 
the  steam,  now  on  the  left  of  the  piston,  and  the  total  atmos- 
pheric pressure  on  the  right  face.  Friction  on  the  guiding  sur- 
faces is  neglected,  and  since  the  motion  is  horizontal  the  weight 
of  the  piston  has  no  component  along  its  path. 

If  P  is  constant  throughout  the  whole  stroke  (left  to  right), 
and  =  6000  Ibs.,  and  the  crank  turns  uniformly  at  the  rate  of 
(u  :=)  200  revolutions  per  minute,  r  being  =  8  inches;  what 
must  be  the  value  of  the  pressure  P1  between  crank-pin  and  the 
side  of  the  slot  just  after  the  dead-point  C  is  passed,  i.e.,  at  the 
beginning  of  the  stroke  ?  Ditto,  when  the  crank-pin  is  45°  from 
C\  and  again,  when  it  is  at  0,  90°  from  (7?  The  weight  of  the 
piston  and  rod  is  160  Ibs. 

Between  C  and  0,  P'  is  smaller  than  P  and  is  a  resistance,  as 
regards  the  motion  of  the  piston.  If  that  motion  were  uniform 
the  full  amount  of  the  6000  Ibs.  would  be  felt  at  the  pin,  for  in 
that  case  the  acceleration  would  be  zero  and  the  horizontal  forces, 
P  and  P',  would  be  equal  and  oppositely  directed  ;  but  from  C 
to  0  the  motion  is  accelerated  (the  piston  has  no  velocity  at  C\ 


HARMONIC    MOTION   OF   PISTON.  69 

so  that  a  certain  amount,  =  mass  X  accel.,  of  the  6000  Ibs.  is 
absorbed  in  the  "  inertia"  of  the  mass,  so  to  speak,  leaving  only 
the  remainder  to  be  felt  as  a  pressure  P'  at  the  crank-pin.  This 
is  expressed  analytically  by  the  relation  2  (hor.  comps.)  —  Mp ; 
i.e.,  P-P'  =  Mp. 

Beyond  O  toward  D  the  constraint  of  the  mechanism  is  such 
as  to  bring  about  the  gradual  stopping  of  the  piston,  which  at  O 
has  its  greatest  velocity  (—  c,  =  to  that  which  the  pin  has  at  all 
times),  so  that  independently  of  the  6000  Ibs.  on  the  left  the  piston 
is,  as  it  were,  thrown  against  the  crank-pin,  the  pressure  produced 
against  which  at  any  instant  from  0  to  D  must  =  Mp  over  and 
above  the  6000  Ibs.  due  to  steam  action  ;  i.e.,  Pf  =  6000  Ibs. 
-j-  Mp  (where  p  is  the  numerical  value  of  the  acceleration,  whose 
algebraic  value  is  now  negative),  or,  analytically,  P  —  P'  =  Mp 
where  p  has  its  algebraic  value  (negative  when  a  number  is  in- 
serted for  it). 

The  linear  velocity  of  the  crank-pin  is  =  %7tru,  =  2  X  -f  -  f  •  -VA 
=  13.97  ft.  per  sec.  (using  the  ft.,  lb.,  and  sec.).  As  the  pin  ap- 
proaches and  passes  a  dead-point  the  motion  of  the  foot  of  the 
perpendicular  let  fall  from  it  upon  the  horizontal  diameter,  along 
that  diameter,  is  not  only  the  motion  of  the  piston,  but  is  at  this 
point  normal  to  the  path  of  the  pin ;  hence  the  normal  accelera- 
tion of  the  pin  is  the  actual  acceleration  of  the  piston  at  a  dead- 
point.  Therefore  c2  -f-  r  (see  eq.  (4),  p.  75 ;  and  also  section  §  75, 
and  first  line  of  p.  60,  M.  of  E.)  is  the  acceleration  of  the  piston 
at  (7;  and  therefore  just  after  passing  the  dead-point  O  we 
have 


—  6000  —  1452  =  4548  Ibs. 


The  acceleration  of  the  piston  is  proportional  to  the  displace- 
ment, and  hence  at  45°  from  C  we  have 

P'  =  P  —M\G-  cos  45°J  =  6000  —  1452  X  .707  =  4937  Ibs. 
At  0,  p  =  0  and  P  =  P'  =  6000  Ibs. 


70  NOTES   AND    EXAMPLES   IN   MECHANICS. 

Beyond  0  at  45° 

P'  =  6000  + 1452  X  .707  =  7063  Ibs. ; 
while  just  before  reaching  D 

P'  =  6000  +  1452  =  7452  Ibs. 

On  the  return-stroke  steam  is  admitted  to  the  right  of  the  piston 
and  P'  occurs  on  the  other  side  of  the  slot,  with  same  variation 
during  the  stroke  as  before. 

67.  Example  15.     Conical  Pendulum,  or  Simple  Governor-ball 
(Fig.  82  [a]).— If  the  oblique  part  of  the  cord  is  to  be  20  in.  in 

length,   what   tangential  velocity  in 
a  horizontal  circle  (centre  at  C)  and 
what  radius  =  ^,  for  that  circle,  must 
be  given  to  the  material  point  G  of 
10  Ibs.  weight  (=  G)  in  order  that 
motion   in   the   circle   shall  be  self- 
[a]       (TJcUoK*.       [bj         perpetuating  and  the  weight  G'  of 
FIG.  82.  40  Ibs.    may  be   sustained  at   restl 

Fig.  82  [b~\  shows  the  moving  weight  as  a  free  body,  the  only 
forces  acting  being  a  gravity-force  of  10  Ibs.  and  an  oblique  cord- 
tension,  P,  which  by  above  conditions  is  to  be  40  Ibs.  The  mo- 
tion of  G  being  confined  to  a  horizontal  plane,  it  has  no  vertical 
acceleration  ;  therefore  2  (vert,  compons.)  should  balance,  or 

p  cos  OL  —  G  =  0,  whence  cos  a  =  -p  =  0.25,  and  a  should 

=  75°  31/.  Hence  r  should  be  made  =  (20  in.)  x  sin  a  =  20 
X  0.968  —  19.36  in.  =  1.613  ft.  Since  the  motion  is  to  be  in  a 
curve,  2  (normal  compons.)  at  any  instant  should  —  M  X  (vel.)* 

G    tf 
-r-  rod. ;  i.e.,  P  sin  a  -[-  0  =  —  •  —  ;  and  combining  this  with  the 

\J 

P  cos  a  =  G  derived  above  we  have  tan  a  =  c*  -+-  gr,  whence 
the  required  velocity  must  be  c  =  1/32.2  X  1.613  X  3.871  =  14.2 
ft.  per  sec.  The  proper  radius  being  as  above  (r  =  1.613  ft.), 
this  implies  rotation  about  C  at  the  rate  of 

c  14.2 

=  1.40  revolutions  per  second, 


or  84  per  minute. 


CONICAL   PENDULUM — EXAMPLE.  71 

68.  Example  16.  The  Weighted  Governor  or  Conical  Pendulum 
(Fig.  83). — The  four  inextensible  cords,  each  16  inches  long, 
connect  the  three  "material  points,"  or  small  „/„/,/„„$,//„„„„* 
bodies,  as  shown  ;  the  two  upper  cords  being  at- 
tached to  a  fixed  support  at  A.  Gl  and  G,  are 
two  balls  of  equal  weight;  the  weight  of  each 
=  Gl  =  8  Ibs.,  while  the  block  #2  weighs  12 
Ibs.  If  now  the  balls  are  caused  to  rotate  about 
the  vertical  axis  A  CB  at  slowly  increasing  rate 
(revolutions  per  minute),  by  pressing  against  them  FlG  83- 

laterally  with  a  vertical  board  whose  plane  contains  the  axis 
ACB,  the  angle  a  gradually  increases  and  the  block  6r2  is  lifted 
along  the  axis  towards  A.  When  the  speed  of  rotation  has 
reached  any  desired  figure  (rev.  per  min.)  a  has  some  correspond- 
ing value,  and  if  the  board  is  now  removed  a  retains  that  value 
and  the  balls  continue  their  motion  (forever,  if  no  friction)  in  the 
corresponding  horizontal  circle,  sustaining  the  block  6r2  at  rest 
(at  least  its  centre  of  mass  is  at  rest)  in  some  position  B. 

Required  the  distance  AB,  —  2A<7,  when  a  speed  of  rotation 
of  120  revs,  per  min.  has  been  attained  ? 

Let  I  denote  the  cord-length  of  16  in.,  r  the  unknown  radius 
of  the  horizontal  circle,  Sl  the  tension  induced  in 
each  of  the  upper  cords,  £2  that  in  the  lower,  and  c 
the  unknown  linear  velocity  of  each  ball.  Let  u  = 
the  number  of  revolutions  per  unit  time  (so  that 
c  =  %7rru).  Fig.  84  shows  one  of  the  balls  as  a  free 
body.  Since  its  vertical  velocity  is  always  the  same 
(zero),  i.e.,  its  vertical  acceleration  =  0  (the  motion 
being  confined  to  a  horizontal  plane), 

2  (vert,  comps.)  =  0 ;  i.e.,  £,  cos  a  —  $,  cos  a  —  Gl  =  0 ;     .     (1) 

while  on  account  of  the  curvilinear  motion  "2  (normal  compons.) 

=  Me*  -f-  r,  or 

in  or  =  —  •  - (2) 


72  NOTES   AND   EXAMPLES   IN  MECHANICS. 

The  tangent  to  the  curve  is  -|  to  the  paper  and  2  (tang,  comps.)  is 
evidently  =  0,  whence  the  tangential  acceleration 
must  be  zero  ;  i.e.,  c  is  constant,  as  we  have  as- 
sumed all  along. 

With  6r2  free,  in  Fig.  85,  we  have  balanced 
forces  ;  whence  2  (vert,  compons.)  =  0;  i.e., 

2$,  cos  a  —  £2  =  0 (3) 

By  elimination,  noting  that  c  = 


2)  tan  a  r  - —  q         G.-4-  G9 

*i .    i  p      —     A  ( 1    y  'I          a 

' /'"y  «      J«ty«»      ,  *      .JllL  \J  +      ..         oo  /-*  '   • 

ffr  6r,  '  tan  a?  k.rfu  Gl 

We  note,  therefore,  that  the  required  distance  AC  is  inde- 
pendent of  the  length  I  and  is  inversely  proportional  to  the  square 
of  the  number  of  revolutions  per  unit  time.  (A  similar  result 
was  found  with  the  simple  conical  pendulum  ;  see  p.  78,  M. 
of  E.) 

Hence,  numerically,  with  the  foot,  pound,  and  second  (so  that 
u  =  -1//  =  2  revs,  per  sec.), 

2  v  32  2        g  i   10 

AB,  =  2 AC,  =  4  x  987*x4 '       8 —  =  L018  ft< ;  or  12'22  in' 
69.  Example  17.    Cannon-ball  under  Gravity  and  Air-resistance. 
— A   round   cannon-ball,    weighing    24 
Ibs.,  is  at  a  certain   point  of  its  path 
moving  with  a  velocity  of  v  =  800  ft. 
t~b~ —  per  second  in  a  direction   making   an 
angle  of  20°  below  the  horizontal.    The 
resistance  offered  to  it  by  the  air  at  this 
Fl0-86-  speed  is  80  Ibs.  and  acts  in  the  line  of 

motion,  since  the  body  is  round  and  has  no  motion  of  rotation. 
Required  the  amount  and  position  of  the  (ideal)  resultant  force 
E.  See  Fig.  86. 

Since  there  are  only  two  forces  acting  on  the  ball,  P  and  G, 
E  must  have  an  amount  and  position  determined  by  the  diagonal 
of  the  parallelogram  formed  on  P  and  G.     See  figure  for  the 
known  angles.     Hence  (from  formula  on  p.  7,  M.  of  E.) 
E  =  VP*  +  CT  +  2PG  cos  110°,     i.e., 
E  =  V80a  +  242  +  2  x  80  X  24  X  (—  0.3420)  =  75.25  Ibs. 


NUMERICAL  EXAMPLES— DYNAMICS  OF  MATERIAL  POINT.  73 

To  find  the  angle  0,  note  that  in  the  triangle  PRO  we  have 

24 
sin  6  :  sin  70°  ::  G  :  R\  whence  sin  0  =  =^-^(0.9397);  i.e.,  6  = 

<  O.-ZO 

17°  26',  and  hence  R  is  2°  34'  above  the  horizontal. 

Since  the  action- line  of  R  is  not  coincident  with  the  line  of 
motion  of  the  ball  at  this  instant,  the  path  of  the  hall  must  he 
curved,  the  radius  of  curvature  at  this  point  depending  on  the 
mass,  on  the  square  of  the  velocity,  and  on  the  value  of  the 
normal  component  of  R ;  while  the  rate  of  retardation  of  the 
velocity  (negative  tangential  acceleration)  depends  on  the  mass 
and  the  tangential  component  of  R.  (See  next  example.) 

70.  Example  18.  Ball  in  Curved  Path.  Radius  of  Curvature, 
etc.  (Fig.  87).— A  large  ball  weighing  200 
Ibs.  (so  that  its  mass  =  M  —  200  -f-  32.2 
—  200  X  0.031  =  6.2,  in  the  foot-pound- 
second  system  of  units)  at  a  certain  point 
of  its  path  has  a  velocity  of  700  ft.  per 
sec.,  the  resultant  force  R  at  this  instant 
being  ==  300  Ibs.  and  making  an  angle  of 
140°  with  the  direction  (see  v  in  figure)  of  motion. 

Required  the  radius  of  curvature,  r,  at  this  point  of  the  path, 
and  also  the  tangential  acceleration. 

By  a  rectangular  parallelogram  of  forces  we  resolve  R  along 
the  tangent  and  normal,  obtaining  for  its  tang,  compon., 

T,=R  cos  140°,  =  300  X  (-  0.76604)  =  -  229.812  Ibs., 
while  Ny=B  sin  140°,  =  300  x  0.6428  =  192.84  Ibs. 

From  eq.  (5),  p.  76,  M.  of  E.,  2  (norm,  comps.)  =  Mv*  ~-  r, 
and  2  (tang,  comps.)  =  Mpt ,  whence 

_  6.2  X  (700)-  ...mgof 
192.84  5>T5(  "•» 

—  229.812 
and  pt  = g-g—      =  —  37.06ft.  per  sec.  per  sec. 

This  last  value  means  that  the  retarding  effect  of  the  com- 
ponent Tis  such  that  if  the  rate  of  retardation  remained  constant 


74 


NOTES   AND   EXAMPLES   IN   MECHANICS. 


for  one  second,  at  the  close  of  that  second  the  velocity  in  the 
path  would  be  700  —   37.06  =  662.94  ft.  per  sec. 

71.  Example  19.  Steam  Working  Expansively  and  Raising  a 
Weight.— In  Prob.  4  of  p.  61,  M.  of 
E.,  supposing  the  boiler-gauge  to  read 
80  Ibs.  per  sq.  in.  (above  one  atmos- 
phere)  and  the  total  length  of  stroke. 


f-AH-G-> 

--HT-IR 


FIG.  8?a. 


K  sn  —  ON,  to  be  16  inches,  with  cut- 
off at  one  third  stroke  (so  that  5,  =  -J 
of  16  in.),  the  diameter  of  piston 
being  10  inches  ;  how  great  a  weight 
(}  can  be  raised  if  the  (circular)  pis- 
ton is  to  come  to  rest  at  the  end  of  the  stroke,  having  started 
from  rest  at  the  beginning  of  the  stroke?  Required  also  the 
time  occupied  from  0  to  B,  and  the  position  of  the  piston  when 
its  velocity  is  a  maximum. 

From  p.  62  we  have  the  equation 


.     .     •     (2) 


now  to  be  solved  for  G.  sn  =  f  ft.  ;  sl  =  f  ft.,  and  hence  the 
ratio  sn  :  s,  —  3.  The  area  of  piston  =  xr*  —  %f-  X  (5)2  =  78.57 
sq.  in. 

.-.  Air-pressure  above  piston,  =  A,  =  constant  —  78.57  X  (15 
Ibs.  per  sq.  in.)  =  1178  Ibs.  ;  while  the  steam-  pressure  under  pis- 
ton while  it  is  passing  from  0  to  B,  —  8,  ,  =  78.57  X  95  = 
7464.15  Ibs.  Noting  that  loge  =  common  log  X  2.302,  we  have 
from  eq.  (2)  (using  the  foot,  pound,  and  second) 

7464  X  t[l  +  2.302  X  .47712]  =  1178  X  f  +  G  X  f 

Solving,  G  =  4044  Ibs.  (so  that  M  =  4044  x  .031  =  125.36). 

The  acceleration  from  O  to  B  is  constant  and  — 
pi  —  (Sl  —  A  —  G)-^M  =  2241  -f-  125.36  =  17.88  ft.  per  sec.  per 
sec.  ;  and  [eq.  (2),  p.  54,  M.  of  E.]  st  =  %pj?  ;  hence 


time  from  OtoB 


=  tl  =  y  ^| 


=  V.0496  =  0.222  sec. 


BALL   FALLING   ON   SPRING.  75 

Above  B,  the  steam  -pressure  S  diminishes,  and  when  at  some 
point  m  it  has  become  =  A  +  #,  i.e.,  to  5222  Ibs.,  the  resultant 
or  accelerating  force,  8  —  (A  +  6r),  is  zero  ;  above  this  point  m 
that  force  is  negative,  i.e.,  the  velocity  diminishes,  and  hence  the 
velocity  is  a  maximum  at  m.  Let  sm  be  the  distance  of  m  from 
Oy  then  from  Boyle's  Law  s,  :  sm:  :  5222  Ibs.  :  St  ,  whence 
sm  =  I-  Jfft  =  0.635  feet,  =  T.620  in. 

72.  Example  20.  Ball  Falling  on  Spring  (Fig.  88).—  A  ball 
weighing  two  pounds  (G)  falls  freely  from  rest, 
and  after  falling  5  ft.  (=  h)  comes  in  contact  with 
the  head  of  a  spring,  which  it  gradually  compresses 
during  its  further  descent  until  brought  to  rest 
again  momentarily  (at  w,).  The  resistance  (P)  of 
the  spring  is  proportional  to  the  depth  of  compres- 
sion  (s)  and  is  60  Ibs.  (P0)  at  the  end  of  the  first 
inch  (s0).  (Provision  is  made  against  side-buck- 
ling.)  Required  the  maximum  compression  mjnl9 
=  8,  . 

At  the  end  of  the  free  fall  the  velocity  of  the 
balls  is  c  —  VZyli  (i.e.,  c2  =  2^A),  since  so  far  there  is 
but  one  force  acting,  its  own  weight  G.  At  any 
distance  s,  however,  below  ?n0  (the  point  of  first 
contact)  the  resultant  downward  force  is  G  —  P,  FIG.  88." 
P  being  the  upward  pressure  of  the  head  of  the  spring  against 
the  ball  at  this  instant,  and  the  acceleration  is  therefore  variable 
and  is  p  =  force  ~-  mass,  =  (G  —  P)  -f-  (G  -f-  g).  Let  down  be 
positive.  Substituting  in  vdv  =  pds,  noting  that  P:P0::s:so9 
and  then  integrating  between  the  points  m0  and  m,  ,  we  have 

i  i  i  r       /**>     p   rsi 

—vdv  =  d  s  —  -rfPds  ;       and  /  vdv  =    I  d  s  —  -^-    I  sds. 

g  G  g  /  J  GsQJ 

(,,_o)-g;K:;]1.r,_;=,-^.< 

Numerically,  with  the  inch,  pound,  and  second, 


Finally,  m,ml  ,  =  «,,=-{-  2.03  inches  (and  —  1.96  in.). 


76  NOTES   AND    EXAMPLES   IN   MECHANICS. 

The  negative  result  refers  to  a  point  (call  it  m')  1.96  in.  above 
m0 .  This  is  the  position  where  the  ball  would  momentarily  come 
to  rest  for  the  second  time,  if  it  adhered  to  the  head  of  the  spring 
after  the  latter  had  regained  its  natural  length,  supposing  the 
lower  end  of  the  spring  to  be  fixed.  This  motion  of  the  ball 
while  in  contact  with  the  spring  is  really  harmonic,  whose  central 
point,  from  which  the  u  displacement "  would  be  reckoned,  is  -£$ 
of  an  inch  below  m0 ,  i.e.,  at  the  point  where  the  pressure  of  the 
spring  =  2  Ibs.  (the  weight  of  ball),  so  that  as  the  ball  passes  that 
point  its  acceleration  is  zero  and  the  velocity  a  maximum.  This 
point  is  midway  between  m,  and  in'. 

72a.  The  Engineer's  "  Mass," — The  engineer  measures  the  mass 
of  a  body  (in  case  a  problem  connected  with  its  motion  is  under 
treatment)  by  the  fraction,  weight  -j-  accel.  of  gravity  ;  or  G  -f-  g. 
This  is  not  scientific,  but  is  so  firmly  rooted  in  engineering  prac- 
tice that  no  different  measure  can  well  supplant  it.  It  seems 
to  imply  that  the  amount  of  matter  in  a  body  depends  on  the 
existence  of  the  attraction  of  gravitation;  whereas,  of  course, 
such  is  not  the  case.  This  measure  arises  from  the  fact  that  a 
convenient  way  for  the  engineer  to  determine  the  magnitude  of 
any  force  P  (or  resultant)  acting  on  a  body  and  producing  an 
acceleration  (p)  of  its  velocity  is  to  compare  it  with  the  force  of 
gravity  exerted  on  the  body,  whether  the  circumstances  of  the 
problem  are  affected  by  gravitation  or  not.  In  the  phrase  force 
=  mass  X  accel.,  or  P  =  Mp,  the  word  mass  is  simply  a  name 
given  to  the  fraction  G  -±  g,  the  origin  of  which  is  as  follows : 

In  the  actual  problem  the  force  P  produces  an  acceleration 
=  p  in  the  velocity  of  the  body.  In  the  ideal  experiment  of 
allowing  the  same  body  to  have  a  free  fall  in  vacua  we  know 
that  the  only  force  would  ~be  the  weight  6r,  and  that  the  resulting 
acceleration  would  be  g ;  and  since  the  forces  must  be  propor- 
tional to  the  accelerations,  we  have  (§  54,  M.  of  E.) 

P  :  G  : :  actual  p  :  ideal  g\    or,    P  —  —p. 

In  other  words,  the  engineer  uses  the  gravitation  measure  of  a 
force  (p.  48,  M.  of  E.). 


CHAPTER  Y. 
MOMENT  OF  INERTIA  OF  PLANE  FIGURES. 

72b.  Phraseology. — Unless  otherwise  specified,  we  are  to  under- 
stand by  "  moment  of  inertia  of  a  plane  figure"  the  rectangular 
moment  of  inertia  ;  i.e.,  the  axis  of  reference  lies  in  the  plane  of 
the  figure  (and  not  i  to  it  as  with  the  "  polar"  moment  of  in- 
ertia). This  is  a  useful  function  of  the  plane  figure,  to  be  used 
in  the  theory  of  beams  under  bending  strain. 

73.  Moment  of  Inertia  of  Section  of  I-beam  (Corners  not  Rounded). 
— Fig.  89  shows  the  form  and  dimensions  of 
the  section,  which  is  symmetrical  about  each 
of  the  axes  X  and  J",  and  is  for  present  pur- 
poses subdivided  into  three  rectangles  and 
four  right  triangles.  Making  use,  then,  of  ]  i, 

results  obtained  for  those  elementary  forms,  •*-' ^4~ 

and  of  the  transferral  formula  between  the 
gravity  axis  of  any  figure  and  a  parallel  axis 
(see  p.  94  and  eq.  (4),  p.  93,  of  M.  of  E.), 
we  have  for  the  moment  of  inertia  about 
axis  X 


\  \ 
*l 


FIG.  89. 


-v 


That  is,  numerically, 


=  566  +  195.8  +  356  =  1117.8  bi.  in.  = 
Similarly,  the  moment  of  inertia  about  the  axis  Y  is 


=  22.5  +  0.17  +  9.79  =  32.46  bi-quad.  in.  =  Iy 


77 


78 


NOTES  AND   EXAMPLES  IN  MECHANICS. 


74.  Moment  of  Inertia  of  a  Section  of  a  Built  Box-beam  (Fig.  90). 

— The  beam  is  composed  of  two 
"flange-plates"  (upper  and  lower), 
two  vertical  "stem-plates"  and 
four  "angle-bars"  of  equal  legs, 
riveted  together.  See  figure  for 
notation  and  dimensions.  He- 
quired  the  moment  of  inertia 
of  the  whole  section  about  JT,  its 
horizontal  gravity  axis  of  sym- 
metry. 

In  Fig.  91  we  have  the  loca- 
tion of  the  gravity  axis  g  (parallel  to  X)  of  a  single  "  angle" 
section,  according  to  the  hand-book  of  the  New 
Jersey  Steel  and  Iron  Co.,  so  that  from  the  dis- 
tance 1.68  in.  we  compute  the  d'  =  10.32  in.  of 
Fig.  90,  or  distance  of  axis  g  from  axis  X. 


ar »-*>- =-* 


Allplates  X  thick 
FIG.  90. 


From  the  same  book  we  find  that  the  Ig  of  the 


j 


Section  of 

Angle-Bar 

6X6X.K" 


FIG.  91. 


angle-section  is  20  bi.  in.  (very  nearly),  and  its 

area  F'  =  5.75  sq.  in.     Let  t  =  thickness  of  all  plates  =  £  in.  ; 

and  t'  =  diam.  of  rivet-holes  =  f  in. 

First,  neglecting  the  rectangular  gaps  made  by  the  rivet-holes, 
we  have  the  IJs  of  the  various  component  sections  as  follows  : 

(four  "  angles")  .  .  .  4[/0  +  ^^/2)]=4[20+5f(10.32)2]      =2528  ; 
(flange-plates) 


(stem-plates)  .....  2^] 


=11*2  ; 


making  a  total  of  6681  bi.  in. 

Treating  the  small  rectangles  left  by  the  rivet-holes  as  con- 
centrated in  their  respective  centres  of  gravity  [and  thus  neglect- 
ing their  local  (gravity)  moments  of  inertia], 

"V~ 
Subtractive)  .,,        ' 

Ix  due  to    V  =4[(20-^"a]+4[j204-2J    \  =  243  +  432=675. 
rivet-holes  ) 

Hence,  lx  of  actual  section  =  6681  —  675  =  6006  bi.  in. 


MOMENT  OF  INERTIA  BY  SIMPSON'S  RULE. 


79 


It  will  be  noticed  that  the  first  term,  -^bt3  [or  local  (gravity) 

moment  of  inertia],  in  the  Ix  of  the  section  of  a  flange-plate, 

above,  is  very  small  compared  with  the  second,  or  "  transferral 

term"  (bt  .  d*).     This  is  due  to  the  fact  that  all  parts  of  a  thin 

flange-plate  section  are  very  nearly  at  the  same  distance  from  the 

final   axis  of   reference,  X.     In  such  a  case  it  is  customary  to 

neglect  the   local  term,  as   no   practical   error  results  from  so 

doing. 

75.  Moment  of  Inertia  of  Irregular  Curve-bounded  Plane  Figures 

by  Simpson's  Rule  (see  §  93,  M.  of  E.).  —  If  an  exact  result  for  the 
Ix  of  the  figure  shown  in  Fig.  92  were  desired, 
we  might  first  conceive  of  its  subdivision  into 
narrow  strips  parallel  to  JT,  of  variable  length 
v  and  infinitesimal  width  dz,  then  express  its  lx 
as  f  (small  area)  X  32,  mf(vdz)z*,  =f(z*v)dz  ; 
and  finally  perform  the  integration,  if  v  were 

an  algebraic  function  of  z.     If  such  is  not  the  case,  however, 

(or  if  such  is  the  case  but  the  in- 

tegration  net  practicable,)  we  can 

resort  to  Simpson's   Rule  (§  15), 

noting  that  the  u9  x,  and  dx  of 

that  rule  correspond  to  the  (z*v\ 

z,   and   dz,   respectively,   of    the 

present  problem.     We  divide  the 

whole  height  of  the  figure  (from 

the  axis  X]  into  an  even  number 

n  of  equal  parts,  and  through  each 

point  of  division  draw  a  parallel 

to  X,  thus    determining  a  series 

of  widths,   v0,   vl9  v9,  etc.     For 

example,  this   construction  being 

made  for  the  upper  part  of  the  rail-section  in  Fig.  93,  with  n~ 

we  have  its  Ia9  =f(z*v)dz,  approximately  = 


eths  of  an  inch. 


FIG.  93. 


With  numerical  substitution,  therefore,  the  lengths  marked  in 
the  figure  having  been  scaled  in  fiftieths  of  an  inch  (noting  that 


80  NOTES  AND   EXAMPLES  IN  MECHANICS. 

z0  =  0,  s,  =  %z6  ,  z^  =  f  26  ,  s3  =  f-26  ,  etc.),  we  have,  as  the  Ix  of 
the  portion  of  the  rail-section  lying  above  the  axis  X, 

[0  +  4(1»  X  40  -f  32  X  80  +  5*  X  163)  -f  2(22  X  50  +  42  X  136)  +  6»  X  141] 
bi-quad.  fiftieths  ;  which  divided  by  (50)4,  or  6,250,000,  gives 
Ix  for  upper  part  =  25.27  bi.  in. 

Similarly,  the  vertical  height  of  the  lower  part  being  divided 
into  four  equal  lengths,  we  have  for  the  lx  of  that  part  (nearly) 

SIT?  [°  +  ^  X  53  +  32  X  244)  +  2(22  X  103)  +  42  X  266] 
=  129,870,000  bi.  fiftieths.  Dividing  by  (50)4,  we  have  20.77 
bi.  in.  ;  so  that  the  total  Ix  of  the  complete  rail-section  =  20.77 
+  25.27  =  46.04  bi.  in. 

Xis  a  gravity  axis  parallel  to  the  base  of  the  rail-section,  and 
has  been  located  by  cutting  out  the  shape  from  card  -board  and 
balancing  on  a  needle-point. 

If  the  plane  figure  is  of  such  a  form  that  a  division  into  strips 
perpendicular  to,  and  all  terminating  in,  the  axis 
of  reference  (X)  is  convenient,  the  exact  cal- 
culus form  for  its  Ix  is  %fy*dx  (see  latter  part  of 
§  93,  M.  of  E.),  for  each  strip  is  an  elementary 
rectangle.  See  Fig.  94. 

"o     £x  If  Simpson's  Eule  is  to  be  applied,  divide  the 

FIG.  94.  base  OX  of  the  figure  into  an  even  number,  n,  of 
equal  parts  and  scale  the  extreme  ordinates  y0=  AO  and  yn=  BX, 
also  the  intermediate  ordinates  yl  ,  ya  ,  etc.,  at  the  points  of  divi- 
sion. "With  Ti  —  6,  for  example,  we  have  as  an  approximation 

TTx 
Ix  =  ~ 


76.  Graphical  Method*  for  the  Gravity  Axis  and  Moment  of 
Inertia  of  a  Plane  Figure  (Fig.  $&).—A"B"C"  is  the  figure  in 
question  (drawn  in  full  size).  It  is  required  to  construct  the 
special  gravity  axis,  7?,  that  is  parallel  to  the  base  A"B"^  and 
to  obtain  the  moment  of  inertia  about  that  axis. 

Divide  the  figure  into  strips  parallel  to  A"  '  B'  ',  of  small  width 
*  From  Ott's  Graphical  Statics. 


MOMENT   OF  INERTIA — GRAPHICAL   METHOD. 


81 


(no  width  being  more  than  one  eighth,  say,  of  the  total  width  -|  to 
A"B"\     These  widths  need  not  be  equal. 

Through  the  centres  of  gravity  of  the  strips  (1, 2, 3,  etc.)  draw 
indefinite  lines  (1  . .  1',  2  . .  2',  etc.)  ||  to  A" B"  (with  most  strips  it 
is  accurate  enough  to  take  the  centre  of  gravity  midway  between 
the  sides).  Along  any  right  line  0' I,  parallel  to  A"  B" ,  lay  off  the 
lengths  O'A,  AB^  BC,  etc.,  proportional,  respectively,  to  the  areas 

R 


Moment  of 

Inertia 
Graphic  Method. 


FIG.  95. 

FI  ,  Ft ,  etc.,  of  the  successive  strips,  in  the  order  and  position 
shown  (in  most  instances  *  each  such  area  may  be  assumed  pro- 
portional to  the  length  of  the  strip  measured  through  the  centre 
of  gravity).  Through  0'  and  /  draw  lines  at  45°  with  0'7,  as 
shown,  to  determine  the  "pole"  0,  from  which  the  "  rays"  OA, 
OB,  etc.,  are  now  drawn.  Then  through  any  convenient  point 
Z  draw  ZC'"  \\  to  00'.  From  the  intersection  a  with  1 . .  1' 
draw  ab  \\  to  OA  to  intersect  2 .  .  2'  in  some  point  5,  then  be  \\ 
to  OB,  and  so  on  ;  until  finally,  through  i,  ids  drawn  ||  to  IO 
to  determine  C  by  intersecting  ZC'" .  The  required  gravity  axis 
R  passes  through  C  \\  to  A" B" .  (For  proof  consult  §  376,  M. 
of  E.) 

The  moment  of  inertia  about  axis  R  may  be  obtained  by 

multiplying  together  the  area  of  the  given  figure  A" B" C"  (call 

it  F)  hy  the  area  (call  it  Fr)  of  the  "  inertia-figure"  (i.e.,  the  area 

included  between  the  two  lines  Ca  and  (7*,  and  the  broken  line, 

*  When  the  strips  are  narrow  and  of  equal  width. 


82  NOTES  AND   EXAMPLES   IN   MECHANICS. 

or  "equilibrium-polygon,"  dbcdefghi ;  shaded  in  Fig.  95).  The 
proof  of  this  relation  (strictly  true  only  for  infinitely  narrow 
strips)  is  as  follows : 

At  any  vertex  of  the  equilibrium-polygon,  as  at  g,  there  are 
two  segments  meeting;  prolong  them  to  intersect  the  gravity 
axis  R  in  some  two  points,  as  m  and  n.  Then  mng  is  a  triangle 
with  base  m  . .  n  (call  it  &,)  and  altitude  a?7 .  But  on  the  left  of 
Fig.  95  the  shaded  triangle  OF^  is  evidently  similar  to  mng\ 
whence  the  proportion  #7 :  a?7  : :  F,  :  ^F\  i.e.,  F^x,  —  \F~k, . 
Multiplying  by  a?7 ,  we  have  F^x?  =  F(^xjc,). 

Now  FJK*  is  the  moment  of  inertia  (about  It)  of  strip 
No.  7  (considered  infinitely  narrow),  and  -J^,&7  is  the  area  of  the 
triangle  mng.  We  have  therefore  proved  that  the  moment  of 
inertia  of  any  one  strip  is  equal  to  the  product  of  the  whole  area 
F  of  the  given  plane  figure  by  the  area  of  a  triangle  like  mng 
and  obtained  in  a  similar  manner.  If  all  the  triangles  like  mng 
were  drawn,  their  united  areas  would  evidently  be  that  of 
the  "inertia-figure"  abcdefghi-C-a.  Hence  the  sum  of  the 
moments  of  inertia  of  all  the  strips,  i.e.,  the  moment  of  inertia 
of  the  whole  figure  A"B"C",  is 

j     _  j  area  of  plane      \        (       area  of  the 
(figure  A"  £"C"  \-\  "inertia-figure" 

In  practice  these  areas  are  most  conveniently  and  accurately 
obtained  by  means  of  a  planimeter;  otherwise,  subdivision  into 
small  trapezoidal  strips  may  be  resorted  to.  If  the  scale  of  the 
drawing  is  one  half  of  the  actual  size  the  result  must  be  multi- 
plied by  16,  i.e.,  the  fourth  power  of  2;  and  similarly  for  other 
ratios. 

By  the  use  of  a  planimeter  with  this  actual  figure  (Fig.  95) 
the  results  F=  .90  sq.  in.,  and  F'  =  .42  sq.  in.,  have  been  ob- 
tained ;  therefore  IR  =  0.378  bi.  in. 


CHAPTER  YI. 


DYNAMICS  OF  A  RIGID  BODY. 

77.  Example  of  Rotary  Motion.    Axis  Fixed  and  Horizontal 

(Fig.  96). — The  body  AB  consists  of  an  irregular  solid  and  a 
light  drum,  rigidly  connected  and  mounted  on 
a  horizontal  axle  whose  journals  are  2  inches 
(=  2r)  in  diameter,  at  C\  the  radius  of  the  drum 
being  a  =  5  in.  The  weight  of  AB  is  150  Ibs. 
=  Gl ,  and  its  centre  of  gravity  is  situated  in  the 
axis  of  rotation  (hence  Gl  has  no  effect  as  regards 
changing  rotary  velocity,  having  always  a  zero 
FIG.  96.  moment  about  the  axis).  The  weight  G,  =  20 

Ibs.,  is  attached  to  an  inextensible  cord  which  is  wound  on  the 

drum  and  whose  weight  is  neglected.     A   constant  friction  F, 

=  10.5  Ibs.,  acts  at  the  circumference  of  the  journals. 

It  is  required  to  compute  the  radius  of  gyration  Jcc  of  the 

rotating  body  by  the  experiment  of  noting  the  time  ^  occupied 

by  G  in  sinking  a  measured  distance  s, ,   from   rest.     Suppose 

tfj  —  5  sec.  and  s,  =  10  ft. 

At  any  instant  during  this  motion,  the  tension  in  the  cord 

being  £,  we  have  for  the  angular  acceleration  0 

of  AB,  which  is  shown  free  in  Fig.  97,  taking 

moments  about  axis  C, 


while  at  the  same  instant  the  downward  linear 
acceleration  JP,  =  6a,  of  G,  enters  into  the  rela- 
tion involving  the  sum  of  downward  forces  for  G  as  a  free  body, 

viz.: 


/i 

—  S  =  mass.  X  accel.  =  —p. 


(2) 


83 


84  NOTES  AND  EXAMPLES  IN  MECHANICS. 

Solving  for  p,  we  would  find  it  constant,  as  also  £;  hence  G 
has  a  uniformly  accelerated  rectilinear  motion,  and  AB  a  uni- 
formly accelerated  angular  motion.  Therefore  (with  foot,  pound, 
and  second),  since  (from  eq.  (2),  p.  54,  M.  of  E.)  sl  —  kpt*i  and 
hence  p  =  2sl  -=-  t*,  we  have  p  =  20  H-  25  =  0.80  and  is  the  con- 
stant linear  acceleration,  not  only  of  6r,  but  for  any  point  in  the, 
surface  of  the  drum,  so  that  the  angular  acceleration  of  body  AB 
is  0  =  p  -f-  a  =  0.80  -i-  -f%  —  1.92  radians  per  sec.  per  sec.  From 
eq.  (2)  we  now  have  S  =  20  -  (20  X  .80)  -=-  32.2  =  19.504  Ibs. 
as  the  constant  tension  in  the  cord  ;  and  finally,  from  eq.  (1), 

[19.504  X  A  ~  10.5  X  TV]32.2 
*a  1.92  X  150  :  0.812  sq.tt.; 

i.e.,  the  rad.  of  gyration  Jcc  —  0.901  ft.  =  about  10.8  inches. 

78.  Solution  of  Example  of  Compound  Pendulum.     (For  the 
statement  of  the  example  see  p.  121,  M.  of  E.)     Fig.  98.    (A =6", 
r  =  1.2".     Locate   axis   oo   so  that  the   time  of 
o,  oscillation   t'  shall  be  \  sec.)     Referred   to   the 
horizontal-gravity  axis  cc  we  have  for  the  solid 
cone  (from  §  101,  M.  of  E.)  I0  =  -f^M\f  +  JrA2], 


'     sq.  in.     With  the  inch  and  second  g  =  386.4,  and 
FIG.  98.          gt'*  -t-2jt*=  4.889.     Hence,  from  eq.  (1),  p.  120, 
M.  of.  E., 


8  _  4,889  ±  |/23.90  —  1.556  =  +  9.615,     and     +  0.163  in. 

That  is,  with  either  oo  or  o'o'  as  axis  of  suspension,  co  being 
made  =  0.163  in.,  and  co'  =  9.615  in.,  the  duration  of  an  oscilla- 
tion (of  small  amplitude)  will  be  \  sec.  of  time.  As  a  check  we 
note  that  co  +  co'  —  9.778  in.,  which  =  2  X  4.889  —  length  of 
a  simple  pendulum  beating  half-seconds  (see  above  and  also  eq. 
(3),  p.  119,  M.  of  E.). 

79.  Points  of  Maximum  and  Minimum  Angular  Velocities  iu 
Motion  of  Crank-pin  of  a  Steam-engine. — Fig.  99  shows  the  results 
of  applying  the  tentative  graphic  method  mentioned  in  the  mid- 
dle of  p.  124,  M.  of  E.  In  making  trials  for  successive  positions 
of  the  connecting-rod,  advantage  can  be  taken  of  the  fact  that  if 


ROTATION— EFFECT  ON  BEAKINGS. 


85 


the  point  be  found  where  the  axis  of  the  connecting-rod  (pro. 
longed  if  necessary)  intei  sects  the  vertical  line  (vertical  in  this 
instance),  drawn  through  the  centre  C  of  the  crank  circle,  the 
distance  of  that  point  from  C  represents  the  amount  of  the 
tangential  component,  T,  of  the  force  P' ',  on  the  same  scale  on 
which  the  length,  C . .  n,  of  the  crank  would  represent  the  force 
P.  Hence  various  positions  of  the  connecting-rod  are  drawn 


until  those  two  are  found  for  which  the  distance  mentioned  above 
is  f -|,  i.e.,  i^,  of  radius  C . .  n. 

In  the  working  of  most  engines  the  steam  is  used  expansively 
so  that  P  is  variable,  being  greatest  near  the  beginning  of  the 
stroke.  This  would  cause  the  points  n  and  m  to  be  nearer  to  A  ; 
and  similarly,  on  the  return-stroke,  n"  and  m"  nearer  to  B. 

80.  Rapid  Rotation  of  a  Body  on  a  Fixed  Axis.  Effect  on 
Bearings. — So  long  as  a  wheel  or  pulley  in  a  machine  is  perfectly 
symmetrical  about  the  axis  of  rotation  the  pressures  on  the  bear- 
ings are  due  simply  to  the  weight  of  the  pulley  and  the  pulls  or 
thrusts  of  cogs,  belts,  cams,  etc.,  which  may  be  acting  on  the 
pulley  or  on  the  shaft  on  which  it  is  mounted  ;  whether  rotation 
is  proceeding  or  not.  But  if,  through  any  imperfection  in  the 
adjustment  or  mounting,  the  centre  of  gravity  lies  outside  of  the 
axis  of  rotation  (its  distance  from  that  axis  being  p),  other  press- 
ures are  brought  upon  the  bearings,  the  same  in  amount  as  if  the 
pulley  were  at  rest  and  a  force  ==  oo^Mp  acted  through  the  centre 
of  gravity,  away  from,  and  ~j  to,  the  axis  of  rotation. 


86  NOTES   AND   EXAMPLES   IN   MECHANICS. 

For  example,  let  the  pulley  in  Fig.  100,  of  weight  =  644  Ibs., 
be  out  of  centre  by  one  fourth  of  an  inch  and  be 
rotating  uniformly  at  the  rate  of  210  revolutions 
per  minute,  i.e.,  J-  rev.  per  second.  Its  angular 
velocity  is  therefore  GO  =  22  radians  per  second 
and  the  pressures  on  the  bearings  due  to  the  eccen- 

FIG.  100.  tricity  of  the  pulley  at  this  speed  is  the  same  as  if 
the  pulley  were  at  rest  and  a  "  centrifugal  force"  P,  =.  [484(644 
-r-  32.2) -^-g-],  —  201.6  Ibs.,  acted  on  the  body,  as  shown  in  figure, 
where  G  is  the  centre  of  gravity.  The  pressures  on  the  two 
bearings  due  to  this  ideal  "  centrifugal  force"  will  be  inversely 
proportional  to  their  distances  from  the  pulley-centre  and  parallel 
to  P  (aside  from  the  pressures  due  to  the  weight  of  the  pulley, 
action  of  belts,  cogs,  etc.). 

The  continual  change  of  direction  of  these  centrifugal  press- 
ures, as  the  wheel  revolves,  is  likely  to  cause  injurious  vibrations 
in  the  framework  of  the  machine. 

81.  "  Centrifugal  Couple."  (Continuing  the  discussion  of  the 
last  paragraph.) — Even  if  the  axis  of  rotation  does  contain  the 
centre  of  gravity  (or  rather  centre  of  mass  in  this  connection)  of 
the  revolving  body ;  if,  even  then,  we  find  that  any  plane 
containing  the  axis  divides  the  body  into  two  parts,*  the  right  line 
connecting  whose  centres  of  mass  is  not  perpendicular  to  the  axis, 
the  ideal  centrifugal  forces,  Go'2Mlp1  and  oo^M^,  of  these  two 
component  masses  form  a  couple,  causing  pressures  at  the  two 
bearings  ;  which  two  pressures,  themselves,  form  a  couple.  For 
example,  Fig.  101,  where  the  axis  of  the  shaft  AB  and  the  cen- 
tres of  the  two  masses  are  in  the  same 
plane,  and  where  the  products  M^  and 
J/2P2  are  equal,  the  centre  of  mass  of 

the  whole  rigid  body  must  lie  in  the  /'^  \a 

axis  of  rotation  (i.e.,  neglecting  the  mass     M2  y  b  i 

of  the  shaft) ;  but  the  line  joining  the   ' 
centres  of  the  component  masses  is  not 
perpendicular  to  the  axis  AB.     A  and 
B  are  fixed  bearings  and  a  uniform  ro-  ~  FIG.  101. 

tation  with  angular  velocity  of  GO  is  proceeding.     Evidently  the 

*  We  here  suppose  each  of  these  parts  to  be  homogeneous  aud  to  be  sym- 
metrical about  a  plane  passed  through  its  center  of  gravity  aud  perpendicular 
to  the  axis  of  rotation.  If  such  is  not  the  case,  we  must  resort  to  the  principles 
of  £§  \22-l22c  iuclus.  of  M.  of  E.,  to  determine  the  pressures  at  the  bearings. 


PRACTICAL   NOTES   ON   PILE-DRIVING.  87 

action  on  the  bearings  is  the  same  as  if  the  body  AB  were  at 
rest  and  were  acted  on  by  the  two  ideal  centrifugal  forces 
Pl  =  GotMlpl  and  P^  —  &>2J/2/J2,  in  the  positions  shown  in  the 
figure  (on  the  right).  These  forces  are  equal  (since  Mlpl  =  -3/apa), 
forming  therefore  a  couple,  so  that  the  reactions  at  the  two  bear- 
ings (pressures  of  bearings  on  shaft)  would  form  a  couple  of 
moment  P'~b  =  P^a. 

For  example,  let  Ml  weigh  120  Ibs.  ;  J/"a  ,  100  ;  the  distances 
~pl  and  p9  being  2  and  2.4  ft.,  respectively  ;  while  the  angular 
velocity  is  22  radians  per  second  (210  revs,  per  min.).  Also  let 
a  =  3  f  t.  and  I  —  4  ft.  ;  then 


pi  =  f  p*=  -  (22)2    x  2-4  =  2no  ibs- 

So  long  as  the  rotation  is  uniform  these  pressures  P'  lie  in 
the  plane  of  the  axis  and  the  two  mass-centres  ;  which  plane,  of 
course,  is  continually  changing  position. 

82.  Piles.  (See  p.  140,  M.  of  E.)—  As  a  practical  matter  it 
should  be  understood  that  as  the  head  of  the  pile  becomes  broomed, 
or  splintered,  during  the  driving,  the  penetration  occasioned  by 
the  blow  may  be  much  smaller  than  would  otherwise  be  the  case 
(only  one  quarter  as  much  in  some  cases)  ;  hence  it  is  economical 
of  power  that  the  head  be  adzed  off  occasionally,  and  especially 
should  this  be  done  just  previously  to  the  last  few  blows  when 
the  measurement  of  penetration  is  made,  to  be  used  in  a  formula 
for  safe  bearing  load. 

A  formula  for  the  safe  load  has  been  proposed  (see  p.  185, 
Eng.  News,  Feb.,  1891)  in  which,  besides  adopting  the  divisor  6 
of  eq.  (2),  p.  140,  M.  of  E.,  the  value  of  s'  is  assumed  one  inch 
greater  than  that  actually  observed.  In  cases  where  the  actual  s' 
is  small  (as  an  inch  or  less)  this  allows  a  very  wide  margin  of 
safety.  If  s'  is  large,  the  margin  of  safety  is  practically  little 
more  than  that  afforded  by  the  divisor  6.  This  formula  may  be 
written  : 

«   »  7     ,  _  1    _  &h  _        (    or,  for  the   )       1       Gh 

~  6  V  +  one  inch  '      (   Ib.  and  ft.,    i  ~  6   «'  +  TV  ' 


88  NOTES   AND   EXAMPLES   IN   MECHANICS. 

while  if  G  is  in  Ibs.,  h  in  feet,  and  sf  in  inches,  it  takes  the  form 

*    7      J    '      71 

safe  load  ^n  Ibs.  = 


,    ,    -.  . 

(See   Baker's   Masonry   Construction,  and   Trautwine's   Pocket 
Book,  for  many  practical  details  in  the  matter  of  piles.) 

In  experiments  in  the  laboratory  of  the  College  of  Civil 
Engineering  at  Cornell  University,  with  round  oak  stakes  2  in. 
in  diameter  and  a  4-lb.  hammer  falling  four  feet,  the  actual  bear- 
ing loads  have  been  found  to  be  from  about  one  half  to  seven 
eighths  of  that  given  by  the  expression  GJi  -~  s'. 

83.  Kinetic  Energy  of  Rotary  Motion.  Numerical  Example.  — 
The  grindstone  in  Fig.  102  has  an  initial  angular  velocity,  GOO  , 
corresponding  to  180  rev.  per  min.,  about 
its  horizontal  geometrical  axis,  which 
is  fixed.  How  many  turns  will  it  make 
before  it  is  brought  to  rest  by  the  two 
frictions,  at  A  and  at  6"?  The  pressure 
of  the  plank  OD  on  the  stone  at  A  is 
due  to  the  weight  of  180  Ibs.  suspended 
at  rest  at  D.  Assume  that  the  friction, 
F')  or  tangential  action  of  the  plank 
against  the  stone,  is  always  one  third  of 
the  normal  pressure,  which  is  vertical  and 
is  evidently  360  Ibs.  (since  the  plank  is  a  body  at  rest  and  therefore 
under  a  balanced  system  of  forces,  so  that  by  moments  about  O 
we  find  the  normal  pressure  to  be  double  the  180  Ibs.).  Hence 
the  friction  F'  is  =  120  Ibs.,  and  with  regard  to  the  plank  is  a 
force  directed  toward  the  right  ;  but  toward  the  left,  as  regards 
the  revolving  stone.  The  stone  is  a  homogeneous  right  cylinder 
weighing  600  Ibs.,  and  of  radius  =  2',  its  geometrical  axis  being 
the  axis  of  rotation.  Radius  of  journals  at  C  is  1  in.,  and  the 
mass  of  the  projecting  axle  is  neglected.  The  journal  friction, 
F",  tangent  to  the  circumference,  is  to  be  taken  as  constant 
and  as  being  ^V  of  the  normal  pressure,  N"^  on  the  journal. 


FIG.  102. 


WORK    AND   ENERGY   IN    ROTARY   MOTION.  89 

We  now  consider  the  stone  free  in  Fig.  103. 
acting  are  as  shown,  and  the  rotation  clockwise 
with  retarding  angular  velocity. 

Though  the  system  is  not  a  balanced  one  as 
regards  rotation,  since  the  moment-sum  about  C 
is  not  zero  but  =  OMk*,  it  is  such  as  regards 
motion  of  the  whole  mass  vertically ;    for  the 
centre  of  mass  has  a  zero  vertical  acceleration, 
being  at  rest  at  all  times.     Hence  we  may  put 
2  (vert,  comps.)  =  0,  and  thus  obtain,  noting  that  the  pressure 
N"  is  practically  vertical,  N"  =  G  +  N1 ;   and  hence  F"  = 
960  -=-  20  =  48  Ibs. 

To  apply  here  the  principle  of  Work  and  Energy,  as  expressed 
in  eq.  (2)  of  p.  144,  M.  of  E.,  the  range  of  motion  considered 
being  that  occupied  by  the  stone  in  coming  to  rest,  we  note,  Fig. 
103,  that  G,  N',  and  N"  are  neutral;  that  F'  and  F"  (both 
assumed  constant)  are  resistances ;  and  that  there  are  no  working 
forces.  Let  u  denote  the  unknown  number  of  turns  made  in 
coming  to  rest,  ds'  an  element  of  the  path  of  the  point  A,  and 
ds"  an  element  of  the  path  of  a  point  in  circumference  of  the 
journal.  With  the  foot,  pound,  and  second,  we  have,  therefore  : 

Work   )          /*»  /•» 

done    \  =   /     F'ds'  =  F'  /   ds'  =  [120  X  u  X  2*  x  2]  ft.  Ibs.; 

m  F'  )      J0  J0 


on 


Work  )          rn  /*n 

done    \  =    \  F"ds"  =  F"  /  ds"  =   [48  X  u  X  2*  XTV]  ft.  Ibs. 
on  F"  }      J0  Jo 

Since  Jcc*  for  the  cylinder  (see  p.  99,  M.  of  E.)  is  =  %r*  =  2  sq.  ft, 
and  the  initial  angular  velocity  is  G?O  =  27r(180  ~  60)  =  18.85 
radians  per  .second,  we  have 

1  }  =  ^V  -  [«18-85)''£  X  2]  ft.  Ibs. 

The  final  kinetic  energy  is  to  be  zero  and  the  work  of  the  work- 
ing forces  is  zero  ;  hence  from  eq.  (2),  p.  144,  M.  of  E., 

0  =  4807TW  -f  STTU  +  [0  —  (18.85)3(18.6)].    .     .     (3) 
Solving  this,  we  obtain  u  =  4.33  turns,  for  the  stone  to  be 


90  NOTES   AND   EXAMPLES   IN   MECHANICS. 

brought  to  rest.     Eq.  (3)  might  be  read :   the  initial  K.  E.  is 
entirely  absorbed  in  the  work  of  overcoming  friction. 

(How  would  the  time  of  coming  to  rest  be  determined  ?) 
84.  Work  and  Kinetic  Energy.     Motion  of  Rigid  Body  Paral- 
lel to  a  Plane.     Numerical  Example. — Conceive  of  a  rigid  body, 
or  hoop,  Fig.  104,  consisting  of  two  material  points,  M'  and  M ", 

so  connected  with  a  rigid,  but 
imponderable,  framework  that 
the  centre  of  gravity  of  the  two 
material  points  lies  at  the  centre 
of  the  circle  formed  by  the  outer 
edge  of  the  frame.  This  outer 
circle,  or  wire,  is  to  roll  without 
FIG.  104.  slipping  from  a  state  of  rest  (cen- 

tre at  o)  to  a  position  6  ft.  lower,  vertically  (centre  at  n) ;  </,  in 
the  figure,  representing  any  intermediate  position.  The  two 
masses  are  Mr  =  2  and  M"  =  3  in  the  ft.-lb.-sec.  system  of  units, 
their  positions  relatively  to  the  frame  being  shown  in  the  figure. 
Assuming  that  in  passing  position  n  the  two  masses  are  in 
line  with  the  point  of  rolling  contact  m  (Mf  uppermost),  required 
their  respective  linear  velocities,  vn'  and  0n",  at  that  instant.  On 
account  of  the  perfect  rolling,  each  mass  is  at  this  final  instant 
moving  in  a  line  ~j  to  the  line  connecting  it  with  m,  and  vn'  and 
v^'  are  proportional  to  these  distances  ; 

i.e.,  vn'  :  vn"  ::  3  ft.  :  1  ft. ;  or  vn'  =  3vn". 
Throughout  this  range  of  motion  (o  to  n)  the  acting  forces 
are  G,  =  161  Ibs.,  —  the  combined  weight  of  the  material  points 
applied  at  the  centre  of  the  hoop  (see  AB  in  Fig.  104)  and  the 
normal  and  tangential  components,  N  and  P,  of  the  pressure  of 
the  fixed  curved  track  or  guide  against  the  edge  of  the  hoop  at  a. 
When  any  point  of  the  hoop  is  in  contact  with  the  track,  that 
point  becomes  the  point  of  application  of  both  N  and  P,  and  is 
at  rest,  being  the  one  point  of  the  hoop  about  which  all  others 
are  turning  for  the  instant  ("  centre  of  instantaneous  rotation"), 
hence  both  N  and  P  are  neutral  forces.  As  the  centre  of  hoop 
passes  from  q  to  q'  through  an  element  of  its  path,  G  does  the 
work  Gdh ;  hence  the  total  work  done  by  G  is  Gfdh  = 


SINKING   WEIGHT,  KOTATING   BODY   AND   CORD.  91 

161  Ibs.  X  6  ft.  =  966  ft.-lbs.  of  work.     The  initial  K.  E.  of  the 
whole  body  is  =  zero,  and  its  final  K.  E.,  i.e.,  its  (K.  E.)n,  is 


.-.  966  =  i</2[18  +  3]  -  0  ;     or    vn'n  =  92  ; 
whence  vn"  =  9.59  ft.  per  sec.  ;  and  also  vn'  =  28.77  ft.  per  sec. 

[If  a  single  material  point  were  to  slide  from  rest  down  a 
smooth  fixed  guide  through  this  vertical  height  of  6  ft.,  its  final 
velocity  would  be  (  1/2  X  32.2  X  6  =)  19.65  ft.  per  sec.]. 

If  the  hoop  in  the  above  example  were  to  slip  on  any  part  of 
the  guide,  instead  of  rolling  perfectly,  the  force  P  would  no 
longer  be  neutral  ;  and  if  the  edge  of  the  hoop  were  to  indent 
the  surface  without  immediate  and  perfect  recovery  of  form  on 
the  part  of  the  latter,  N  and  P  would  not  be  neutral.  (See  §  172, 
M.  of  E.) 

85.  Numerical  Example.  Work  and  Energy.  Collection  of 
Rigid  Bodies.  [Read  the  ten  lines  under  eq.  (XVI),  p.  149,  M. 
of  E.]  —  Fig.  105  shows  a  body  mounted  on  a  fixed  horizontal 
axis  containing  its  centre  of  gravity,  of 
weight  =  48.3  Ibs.  (hence  its  mass  = 
1.5,  using  ft.,  lb.,  and  sec.),  its  radius  of 
gyration  about  the  axis  being  &  =  3  ft. 
This  body  includes  a  drum,  as  shown, 
from  which  a  light  inextensible  cord 
may  unwind  until  the  point  A  of  drum 
passes  the  position  A'  (the  cord  being  '"FIG.  105? 

firmly  inserted),  thus  allowing  the  10-lb.  weight  to  sink  vertically. 
C  is  a  smooth  fixed  peg  ;  there  is  no  friction  at  the  journals. 
The  figure  shows  the  initial  position  (rest).  The  10-lb.  weight 
begins  to  sink,  and  the  velocities  to  accelerate,  while  the  cord 
simply  unwinds  until  A  reaches  the  unwinding  point  7>,  beyond 
which  the  point  of  insertion  will  follow  the  arc  DA',  the  portion 
of  cord  between  it  and  C  being  straight. 

Beqnired  the  (final)  angular  velocity  Gon  of  the  rotating  body 
when  A  reaches  A. 

Considering  free  this  collection  of  rigid  bodies  (two  masses 
and  cord),  knowing  that  all  mutual  actions  between  them,  if 
normal  to  surfaces,  are  neutral,  and  assuming  no  internal  friction, 


92  NOTES  AND   EXAMPLES   IN  MECHANICS. 

we  note  that  the  only  forces  external  to  the  collection  are  the 
gravity-force  of  48.3  Ibs.  (neutral  because  its  point  of  application 
does  not  move)  ;  the  normal  reactions  of  the  bearings  against  the 
sides  of  the  journals  (these  are  neutral  because  the  path  of  the 
point  of  application  is  always  ~|  to  the  action-line  of  the  force)  ; 
the  pressure  of  the  peg  against  the  side  of  the  cord  (neutral,  for 
the  same  reason  as  that  just  mentioned)  ;  and  the  gravity-force 
of  10  Ibs.,  which  is  the  only  working  force.  The  path  of  the 
point  of  application  of  this  last  force  is  vertical  and  coincident 
with  the  action-line  of  the  force,  and  the  length  of  this  path 
(which  is  equal  to  that  of  its  own  projection  on  the  action-line)  is 
equal  to  the  total  length  of  cord,  s,  that  runs  over  the  peg.  Evi- 
dently s  =  n  x  2  f  t.  =  6.28  ft.  (while  A  is  passing  to  E\ 
increased  by  the  difference  between  the  distance  EC  (considered 
straight  ;  a  is  very  small  in  this  case)  and  the  straight  distance 
A'C\ 


_ 
i.e.,  s  =  6.28  +  [  Vl6a  +  29  —  14]  =  8.406  feet. 

As  both  the  initial  and  the  final  Kinetic  Energy  of  the  10  -lb. 
mass  are  zero,  and  the  initial  K.  E.  of  the  rotating  body  is  zero, 
the  work  10  X  8.406  —  84.06  ft.  -Ibs.,  of  the  one  working  force, 
is  entirely  expended  in  generating  the  final  K.  E.,  ^co^Mk\  of 
the  rotating  body, 

.-.  84.06  =  JcoBf  X  1.5  X  (3)2;     and     oon=  3.53  rad.  per  sec., 
which  corresponds  to  o?B  -r-  2?r  =  0.562  rev.  per  sec. 

As  to  the  tension,  S',  induced  in  the  chord  as  the  10-lb.  weight 
reaches  its  lowest  point  and  begins  to  ascend,  which  is  when  the 
point  of  insertion  of  the  chord  in  the  drum  passes  through  the 
point  A  (in  its  circular  path),  we  note  that  at  that  instant,  since 
the  chord  cannot  stretch,  the  horizontal  acceleration  of  the  inser- 
tion-point, which  is  its  normal  acceleration  at  A  ',  must  be  the 
same*  as  the  vertical  acceleration  of  the  10-lb.  weight,  and  this  is 
upward,  since  the  downward  velocity  is  slackening.  Hence  the 
resultant  force,  S'  —  G,  on  G  at  this  instant  is  upward,  and 
—  (ft  -f.  g)  x  (normal  accel.  of  a  point  moving  with  linear 
velocity  =  osn  X  2  ft.  in  a  curve  whose  rad.  of  curv.  is  2  ft.), 
i.e.,  -  (#  +  g)  x  (<»B  X  2  ft.)2  -T-  2  ft.  =  7.72  Ibs. 
.-.  S'  =  10  +  7.72  =  17.72  Ibs. 

*  Approximate;  the  error  is  less  the  longer  the  distance  A'  C,  compared  with 
OA'  ,  strictly  true  only  when  the  peg  is  at  an  infinite  distance  toward  the  right 
(A  somewhat  similar  approximation  was  made  on  p.  59,  M.  of  E.,  in  consider- 
ing the  connecting-rod  of  a  steam-engine  as  infinitely  long.) 


INTERNAL   FRICTION  —  LOST   WORK.  93 

At  other  parts  of  the  motion  the  cord-tension  is  smaller.    Let 
the  student  trace  the  remainder  of  the  motion. 

86.  Work  of  Internal  Friction.  —  The  student  should  note  well 
that  the  work  spent  on  the  friction  between  two  rubbing  parts  of 
the  collection  of  rigid  bodies  under  consideration  is  formed  by 
multiplying  the  value  of  the  friction   by  the  distance  rubbed 
through  by  one  of  the  parts  on  the  surface  of  the  other,  inde- 
pendently of  the  actual  path  in  space  of  any  point  of  either  body  ; 
for  instance,  in  the  examples  of  §§  144  and  145,  M.  of  E.,  the 
friction  (F"  =  400  Ibs.)  between  the  crank-pin  and  its  bearing 
(in  the  end  of  the  connecting-rod),  the  range  of  motion  of  the 
mechanism  being  that  corresponding  to  a  half-turn  of  the  crank, 
is  multiplied  by  the  length  of  a  half-circumference  of  the  crank- 
pin  itself,  viz.,  ttr"  ,  =  0.314  ft.  (less  than  4  inches),  whereas  the 
centre  of  the  crank-pin  describes  a  distance  of  nr  —  3.14  ft.  at 
the  same  time. 

87.  Locomotive  at  Uniform  Speed.  —  Required   the  necessary 
total  mean  effective  pressure  PQ  in  the  cylinder  of  a  locomotive 
on  a  level  track  to  maintain  constant  the  speed  of  a  train  whose 
resistance,  J?,  at  that  speed  is  given.     This  resistance  is  due 
entirely  to  frictions  of  various  kinds  in  the  train  and  to  the  resist- 
ance of  the  air,  and,  the  track  being  level,  is  equal  to  the  tension 
in  the  draw-bar  immediately  behind  the  locomotive.     Fig.  106 
shows  the  collection  of  rigid 

parts  forming  the  locomotive  \  G        <-2g-> 

alone  and  the  forces  external 
to  them,  all  mutual  frictions 
being  disregarded.  For  sim- 


plicity  consider  that  there  is  '  •'•'•:V:^^V^^-:-;^T^^ 
only  one  cylinder  and  piston  ;  FlG-  106- 

and  take,  as  the  range  of  motion  of  the  parts,  that  corresponding 
to  one  backward  stroke  of  the  piston,  i.e.,  to  a  half-turn  of  the 
driving-wheels.  The  external  forces  are  :  PQ  on  the  piston  ;  an 
equal  PQ  on  the  front  cylinder-head  ;  Y,  Y,  the  pressures  of  the 
rails  against  the  truck-  wheel  s  ;  $,  S,  those  of  the  track  against 
the  driving-wheels,  which  are  supposed  not  to  slip  on  the  track; 
G,  the  weight  of  the  locomotive  ;  and  7?,  the  tension  in  the  draw- 


94  NOTES   AND    EXAMPLES   IN   MECHANICS. 

bar.  Of  these,  the  Ir's  and  the  S's  are  neutral  for  reasons  given 
in  the  example  of  §  84  (perfect  rolling);  G  also  is  neutral,  the 
elements  of  its  path  being  always  ~|  to  the  action-line  of  the  force ; 
R  is  a  resistance,  overcome  through  a  distance  =  nr ;  the  P0  on 
the  front  cylinder-head  is  a  working  force,  its  point  of  application 
describing  a  horizontal  path  of  length  =  nr ;  while  the  PQ  on 
the  piston  is  a  resistance  whose  application-point  moves  forward 
in  absolute  space  a  distance  =  nr  —  %a.  The  kinetic  energy  of 
the  mechanism  being  the  same  at  the  end  as  at  the  beginning  of 
the  stroke,  the  K.  E.  terms  cancel  out  from  the  equation  of  work 
and  K.  E.,  leaving 

P**r  -  PQ(7tr  -  2a)  -  Rnr  =  0  ;     .-.  P0  = 

itt> 

or  P0  =  the  draw-bar  resistance  multiplied  by  the  half-cir- 
cumference of  the  driver  and  divided  by  the  length  of  one 
stroke.  For  example,  if  a  200-ton  train  offers  a  resistance 
of  10  Ibs.  per  ton  (in  the  draw-bar)  at  a  speed  of  20  miles 
per  hour,  then  with  2.5  ft.  radius  for  the  drivers  and  1  ft. 
radius  for  the  crank-pin  circle,  we  have  P0  —  7857  Ibs. ;  i.e.,  with 
two  cylinders,  3928  Ibs.  for  each  piston.  Suppose  each  piston  to 
have  an  area  of  100  sq.  in.,  then  the  average  amount  by  which 
the  steam-pressure  on  one  side  should  exceed  the  atmospheric 
pressure  on  the  other  is  39.28  Ibs.  per  sq.  in.  As  steam  is  ordi- 
narily used  expansively  when  the  train  is  under  full  headway,  this 
means  an  initial  steam-pressure  at  the  beginning  of  stroke  of  per- 
haps 80  or  90  Ibs.  per  sq.  in.  above  the  atmosphere  (as  a  roughly 
approximate  illustration).  Above,  we  have  assumed  the  drivers 
not  to  slip  on  the  rails.  Slipping  will  not  occur  ordinarily  if 
H  is  less  than  about  one  fifth  of  the  total  weight  resting  on  the 
drivers. 

88.  Locomotive  Tractive  Effort  at  Starting.  Track  Level. — 
"When  the  train  is  once  in  motion  the  steam-pressure  on  the 
piston  in  the  latter  part  of  the  stroke  is  much  smaller  than 
the  average,  so  that  the  speed  slackens  temporarily,  the  great 
mass  of  the  train  acting  as  a  fly-wheel.  To  start,  however,  this 
steam-pressure  must  reach  a  certain  minimum  amount  which, we 


MECHANICS   OF   THE   LOCOMOTIVE. 


95 


FIG.  107. 


FIG.  109. 


call  P.  For  instance,  with  one  engine  at  its  dead-centre,  the 
duty  of  starting  devolves  on  the  other  alone,  which  is  then  in  the 
mid-stroke  position  (nearly),  its  crank  being  vertically  over  (or 
tinder)  the  driver-axle,  and  the  horizontal  component  of  the  pull 
on  the  crank-pin  is  =  P.  Fig.  107 
shows  the  driving-wheel  as  a  free 
body,  the  vertical  components  of 
the  acting  forces  being  omitted,  as 
having  no  moments  about  D,  the 
point  of  contact  with  the  track. 
(Steam  is  now  pressing  on  the  left 
face  of  the  piston,  not  shown,  and 
on  the  hinder  (left  end)  end  of  cylinder-head  (see  Fig.  106  for 
direction  of  motion,  etc.).  The  action  of  the  driver  is  that  of  a 
lever  whose  fulcrum  is  at  D  (no  slip).  P'  is  the  horizontal  press- 
ure of  the  locomotive-frame  against  the  driver-axle,  as  induced  by 
the  pull  P\  and  by  moments  about  D  we  havePV  —  .P(r-\-a)'9 

Considering  now  the  locomotive-frame  in 
Fig.  108,  we  find  a  forward  force 
P'  at  A,  the  bearing  of  the 
driver-axle  ;  the  draw-bar  resist- 
ance R,  backward ;  and  a  back- 
ward steam-pressure  =  P  on 
the  hinder  cylinder-head  at  B ; 


FIG.  108. 


whence,  for   equilibrium  (i.e.,  more  strictly,  for   a   very  small 

(r\ 

acceleration),  P'  —  P  =  It,  and  finally  P  —  \~)R,  as  the  neces- 
sary total  effective  steam- pressure  on  the  one  piston  to  start  the 
train  when  the  draw-bar  resistance  is  R  (neglecting  the  resist- 
ances in  the  locomotive  itself).  (No  vertical  forces  shown.) 

Similarly,  if   the  crank-pin  were  vertically  under  the  axle, 
as  in  Fig.  109,  we  obtain  for  the  pressure  at  axle  (induced  by 

P)  the  value  P'  =  P  —  f-j  P,  [since,  from  moments  about  D, 

P'r  =  P(r  —  a)"].  Hence,  as  regards  the  locomotive-frame,  we 
now  find  a  'backward  force  P'  at  the  axle-bearing  ;  but  there  is 


96 


NOTES   AND   EXAMPLES   IN   MECHANICS. 


now  a  forward  steam-pressure,  P,  against  the  front  cylinder- 
head,  so  that  P  —  P'  =  R,  which  after  substitution  gives  us 

P  =  f-7?,  the  same  as  before. 

\aJ 

If  the  wheels  are  not  to  slip,  the  horizontal  action  of  the  rail 
on  the  driver,  viz.,  P",  Figs.  107  and  109,  must  not  exceed  thje 
ultimate  friction,  F,  or  about  one  fifth  the  weight  on  the  drivers. 

But  P"  =  ^P,  so  that  P"  =  R-,  hence  R  should  not  exceed  F. 

89.  Appold    Automatic    Friction-brake,    or    Dynamometer    of 
Absorption.— Fig.  110  shows  the  Appold  friction-brake,  which  is 

automatic  by  reason  of  its 
*'  compensating  lever,"  BC. 
The  brake  is  formed  of  a 
ring  of  wooden  blocks,  con- 
nected by  an  iron  hoop, 
whose  two  ends  are  pivoted 
to  the  lever,  as  shown,  at 

FIG.  no.  FIG.  in.          C.     The  pulley  W  revolves 

within.  If  the  friction  is  insufficient  to  keep  the  weight  G  raised, 
it  sinks  and  the  end  B  of  the  lever  presses  against  the  fixed  stop 
BK,  thus  tightening  the  hoop,  increasing  the  friction,  and  raising 
G  (and  vice  versa  when  the  friction  is  too  great). 

Fig.  Ill  shows  the  Appold  brake  as  a  free  body.  The  normal 
pressures  of  the  pulley  against  the  wooden  blocks  have  no  mo- 
ments about  (7,  while  the  tangential  actions  (i.e.,  the  frictions) 
have  a  common  lever-arm,  =  «,  about  0.  Hence,  by  equilibrium 
of  moments,  Fa  =  Gb  —  PC,  where  F  is  the  sum  of  the  frictions. 
Evidently  the  pressure  P  at  the  end  of  the  lever  should  be 
known,  for  accuracy. 

(See  correspondence  in  the  London  Engineer,  from  Novem- 
ber, 1887,  to  March,  1888.) 

Hence  if  v  is  the  velocity  (ft.  per  sec.  for  instance)  of  a  point 

in  the  pulley-rim,  the  power  absorbed  is  Z  =  Fv  ==  f  —  —  Jv, 

90.  The  "  Carpen tier"  Dynamometer  of  Absorption.    (Exhibited 
in  Paris  in  1880.) — This  is  automatic,  like  the  Appold  brake,  but 


THE    "  CARPENTIER  "   DYNAMOMETER. 


97 


FIG.  112. 


instead  of  automatically  altering  the  tension  in  the  strap  to  keep 
the  weight  "  floating,"  it  changes  the  arc  of 
contact  A  B  until  the  friction  is  so  altered 
(—  F'}  as  to  equilibrate  the  weight  G'  with 
help  of  the  smaller  weight  G.  See  Fig.  112. 
N  is  a  pulley  keyed  upon  the  shaft  of  the 
motor  to  be  tested,  and  therefore  moving 
with  it.  A  weight  G  is  fastened  to  a  strap 
BA  C  against  which  the  pulley  JV^rubs,  but 
the  upper  end  of  this  strap  is  attached  to  a 
block  C,  which  is  a  rigid  part  of  another 
pulley,  or  disk,  J/,  beyond  the  pulley  N. 
The  disk  M  is  loose  on  the  shaft,  the  block 
C  projecting  over  the  face  of  the  pulley  JV, 
attached  to  the  pulley  M  at  D  and  carries  a  weight  G'  which  we 
at  first  assume  to  be  supported  by  a  fixed  platform  &0 .  When 
the  shaft  begins  to  turn  (see  arrow),  and  the  tension  in  CA 
occasioned  by  the  friction  due  to  the  arc  AB  and  weight  G  is 
greater*  than  G',  then  G'  will  begin  to  rise,  the  disk  M  turning. 
But,  as  M  turns,  the  uppermost  point  of  contact  of  the  strap 
AB  on  the  pulley  N  moves  to  the  right ;  i.e.,  the  arc  of  contact 
AB  becomes  smaller,  with  a  consequent  reduction  of  the  friction 
between  N  and  the  strap,  so  that  after  a  little  the  tension  in  CA, 
pulling  on  C,  is  just  sufficient  (—  $')  to  keep  the  weight  G'  at  rest 
at  some  point  k'.  When  this  state  of  equilibrium  is  reached,  we 
have,  by  moments,  for  the  equilibrium  of  the  strap,  Sfr=F'r-\-Gr', 


The  strap  DG'  is 


so 


___ 
and  from  that  of  the  disk,  Sfr  =  G'r'  ;  i.e.,  F'  = 

that  if  v  =  velocity  of  the  pulley-rim,  the  power  =  Z,  =  Fv, 
==          --  G    \v9  (ft.-lbs.  per  second,  for  instance.) 

91.  Numerical  Example  in  Boat-rowing.  —  As  an  illustration  of 
the  relations  of  the  quantities  concerned  in  a  simple,  but  typical, 
case  of  propulsion  on  the  water,  let  us  suppose,  in  the  problem  of 
Fig.  166  of  p.  161,  M.  of  E.,  that  the  distances  from  the  oar-handle, 
A,  and  oar-lock,  (?,  to  the  middle  of  the  blade  are  9  ft.  and  6  ft., 
respectively  ;  and  that  a  pull  of  20  Ibs.  is  exerted  on  A.  The 
pressure  at  the  oar-lock  is  then  30  Ibs.  ;  and  that  of  tne  blade 

*  Or,  rather,  its  moment  greater  than  that  of  G'. 


98  NOTES  AND   EXAMPLES   IN   MECHANICS. 

against  the  water,  10  Ibs.  Hence  the  boat  is  subjected  to  two 
forward  oar-lock  pressures  of  30  Ibs.  each  ;  to  two  backward 
pressures  against  the  foot-rest,  of  20  Ibs.  each  ;  and  to  some  back- 
ward resistance  R  of  the  water  against  the  hull  (R  depending  on 
the  square  of  the  velocity ;  R  =  zero  if  there  is  no  velocity). 

The  difference  between  the  oar-lock  and  foot-rest  pressures  is 
a  forward  force  of  20  Ibs. ;  and  if  the  velocity  of  the  boat  at  the 
beginning  of  the  stroke  is  such  that  R  is  =  20  Ibs.,  the  effect  is 
to  barely  maintain  that  particular  speed  while  the  20  Ibs.  pressure 
is  acting  on  each  oar-handle,  (and  R  remains  constant  also.)  If 
R  is  smaller  than  20  Ibs.  the  velocity  will  be  accelerated,  and  R 
will  increase  ;  if  larger,  the  velocity  diminishes  (and  R  also),  and 
of  course  will  diminish  at  a  much  more  rapid  rate  when  the  oar 
is  lifted  from  the  water. 

For  an  ordinary  small  skiff  (with  pointed  stern  as  well  as  bow) 
containing  one  person,  the  water-resistance  R  is  (roughly)  about 
one  half  pound  for  each  sq.  foot  of  the  wetted  surface  of  the  hull, 
when  the  velocity  is  10  it.  per  sec.  /  for  other  velocities,  as  the 
square  of  the  velocity. 

If  in  above  case  each  oar-handle,  while  under  the  20  Ibs.  press- 
ure, passes  through  a  distance  AE '=  3  ft.,  measured  on  the  boat, 
and  the  blade  slips  backward  in  the  water  an  absolute  distance 
s,  of  6  inches  (say),  =  J-  ft.,  the  absolute  distance  passed 
through  by  the  boat  will  be  (from  the  geometry  of  the  figure) 
5.5  ft.  The  work  spent^oii  siip  is  2  X  10  X  -J-  =  10  ft.-lbs. ;  so 
that,  of  the  work,  2P  .AE=  2  X  20  X  3  =  120  ft.-lbs.,  exerted 
by  the  oar-handle  pressures,  relatively  to  the  boat  (see  fourth  line 
of  p.  161,  M.  of  E.),  110  ft.-lbs.  remain  for  the  work  of  over- 
coming the  resistance,  R,  and  increasing  the  K.  E.  of  the  mass 
of  boat.  R  is  overcome  through  the  distance  s3  —  CD,  —  5.5  ft., 
so  that  if  all  of  the  110  ft.-lbs.  are  spent  on  R  (i.e.,  if  the  velocity 
is  to  be  maintained  constant),  we  must  have  110  =  R  X  5.5,  i.e., 
R  =  20  Ibs.,  as  proved  above.  [In  order  that  R  may  have  this 
value  with  a  small  skiff  the  velocity  must  be  (roughly)  about  11 
or  12  ft.  per  sec.,  at  the  beginning  of  stroke.] 

Note  that  the  absolute  distance  through  which  the  two  20-lb. 
oar-handle  pressures  work  is  5.5  -f-  3  =  8.5  ft.  But,  of  the  abso- 
lute work  done  by  them,  viz.,  2  X  20  X  8.5  =  340  ft.-lbs.,  in  the 


NUMERICAL   EXAMPLE   IN   BOAT-ROWING.  99 

stroke,  the  amount  2  X  20  X  5.5,  =  220  ft.-lbs.,  is  absorbed  in 
overcoming  the  foot-rest  pressures  through  5.5  ft.,  the  remainder 
being  an  amount  2  X  20  X  3,  =  120  ft.-lbs.,  =  2P .  AE]  to  be 
spent  on  the  work  of  slip,  of  liquid-resistance,  and  change  of 
K.  E.  (if  any). 

92.  Remarks  on  the  Examples  of  §  155,  M.  of  E. — In  the  first 
example  the  work  to  be  computed  is  that  done  by  the  tension 
P,  in  the  draw-bar,  considered  as  a  working  force  acting  on 
the  train  behind  the  locomotive.  If  the  velocity  were  uniform, 
a  value  10  X  200  =  2000  Ibs.  would  be  sufficient  for  P;  but  as 
the  velocity  is  to  be  increased,  the  "  inertia"  of  the  train  is  brought 
into  play  and  the  amount  required  is  three  times  as  great  in  this 
instance.  In  Example  2  P  has  the  same  significance. 

As  to  Example  3,  multiplying  the  15,000  Ibs.  resistance  by 
the  speed  reduced  to  ft.  per  sec.  will  give  the  power  in  ft.-lbs. 
per  sec.  Dividing  this  number  by  550,  we  obtain  461  H.  P.  (see 
p.  136,  M.  of  E.). 

In  Example  4  the  resistance  is  greater  than  before,  in  the 
ratio  of  (10)2  to  (15)',  i.e.,  it  is  2£  times  15,000  Ibs.  The  distance 
through  which  it  is  overcome  per  second  being  1J  times  its  former 
value,  we  find  the  power  spent  on  water-resistance  to  be  (in  ft.-lbs. 

per  sec.)  2J  X  1|  X  15000  X  -Q^~QQ- '  wllicl1  divided  b7  55° 
gives  1556  H.  P.  (That  is,  the  respective  powers  are  as  the 
cubes  of  the  speeds.) 

Example  5.  Since  80  per  cent  of  the  power,  Z,  of  the  work- 
ing force  (steam-pressure)  is  to  be  461  H.  P.,  we  write  0.80  X  L 
=  461,  and  obtain  L  =  576  H.  P. 

Example  6.  If  the  thrust  or  pull  of  the  connecting-rod  of  the 
engine  on  the  crank-pin  be  resolved  at  every  point  into  a  tangen- 
tial and  a  normal  component,  77and  N  (Fig. 
113),  we  note  that  T  is  a  working  force  and 
N  neutral.  Hence  at  this  point  in  the  line 
of  transmission  of  power  we  can  ascribe  all 
the  power  to  the  force  T.  T  is  variable,  and 
by  its  average  value,  Tm,  we  mean  a  value  whose  product  by  the 
length  of  the  circumference  described  by  the  crank-pin  shall  be 
the  same  amount  of  work  as  that  actually  done  by  the  variable 


100  NOTES   AND   EXAMPLES   IN   MECHANICS. 

T  per  revolution.  Now  461  H.  P.  means  253,000  ft.-lbs.  of  work 
per  second,  which  divided  by  1.0,  the  number  of  turns  per  sec., 
gives  the  work  done  by  T  in  each  turn.  Hence  Tm%7t  X  1.5 
should  =  253,000;  or  Tm  =  26,890  Ibs. 

Example  7.  Fig.  114.  At  0  we  have  the  sphere  in  its  initial 
position,  the  forces  acting  on  it  being  its 
own  weight  G  (a  resistance),  and  the  two 
components,  ^Vand  T,  of  the  pressure  of 
the  inclined  plane  against  it.  Since  there 
is  no  slipping  (i.e.,  perfect  rolling),  both  N 
'///////////,  and  T  are  neutral  (see  §  84). 

Let  v0  =  initial  velocity  of  the  centre 
FIG.  in.  of  gravity,  and  GOO  the  initial  angular  ve- 

locity. sf  sin  30°  is  the  unknown  height  through  which  the  centre 
will  rise  before  the  velocity  becomes  zero. 

G    v  2 
The  initial  K.  E.  of  translation  is  —  -  ~  ;    and  of  rotation, 


Now  v9  =  ay  and  &03  —  f  /•*  (p.  102).     Hence  by  eq. 
(XY),  p.  147, 


or    ^  =  0.39  ft 


93.  Efficiency  of  a  Wedge.  Block  on  Inclined  Plane.  —  In  the 
numerical  example  3  of  p.  172,  M.  of  E.,  it  is  to  be  noted  that 
the  "  efficiency"  of  the  mechanism,  or  ratio  of  the  work  usefully 
employed  in  overcoming  Q,  to  that  exerted  by  the  one  working 
force  P,  is  57.6  -=-  153.6  =  37.5  per  cent. 

Fig.  115  shows  as  a  free  body  the  block  mentioned  in  prob- 
lem 6,  p.  172,  M.  of  E.  We  are  to  find  the 
force  P  in  the  given  action-line,  such  that  a 
uniform  motion  (equilibrium)  can  be  maintained 
up  the  plane.  The  action  of  the  plane  on  the 
block  is  represented  by  the  normal  pressure  N 
and  the  tangential  action,  or  friction,/^.  [The  FIG.  115. 
student  should  not  assume  thoughtlessly  that  N  must  =  G  cos  /? 
(the  normal  component  of  G),  for  the  unknown  P9  not  being 


ROLLING   RESISTANCE.     ,    ,  101 

parallel  to  the  plane,  has  an  influence  in  determining  N\  here  it 
tends  to  relieve  the  pressure  on  the  plane.],  />%  forces 
balanced,  2X,  and  2Y,  =  0  ; 


whence  P  cos  a  —fN—  G  sin  /3  =  0; 

and  P  sin  tf  -f-  N  —  G  cos  /?  =  0  ; 

£[sin/?+/cos/?] 
and  finally          P=  —  -         .    1  .     —  .......     (1) 

COS  <*  --r  Sin  a 


Problem  7  calls  for  the  value  of  P  if  the  motion  is  to  be 
down  the  plane;  other  things  as  before.  Fig.  o//  ,x 

116  shows  the  change.     The  friction  acts  in  a      Y 
contrary  direction  ;  P  will  be  smaller ;  JV,  larger. 
Solving  as  before,  we  finally  have 
6?[sinyg  -/cos/?] 
cos  a  —f  sin  a 

We  note  here  [eq.  (2)]  that  for  ft  =  <p  =  the  angle  of  friction, 
P  —  0  ;  and  that  if  /is  large  and  ft  small  (<  0),  P  may  be  nega- 
tive, i.e.,  its  direction  may  need  to  be  reversed,  to  aid  the  body 
down  the  plane. 

94.  Work  Absorbed  in  Rolling  Resistance. — With  perfect  roll- 
ing of  a  smooth  rigid  wheel  upon  a  straight,  fixed,  smooth,  and 
incompressible  rail,  the  pressure  R  of  rail  on  wheel  is  a  neutral 
force  as  regards  work  (call  this  "perfect  rolling"} ;  for  its  point 
of  application,  at  the  foot  of  the  perpendicular  let  fall  from  the 
centre  of  wheel  on  the  rail,  is  motionless  so  long  as  it  is  the  point 
of  application.  But  with  an  inelastic,  compressible  rail,  the 
point  of  application  is  a  little  ahead  (distance  =  b)  of  the  foot  of 
that  perpendicular,  and  it  results,  therefore,  that  for  every  ds 
moved  through  by  the  centre  of  the  wheel  the  rail-pressure  is 

overcome  through  the  small  distance  I— ids ;  so  that  when  the 

wheel-centre  (or  any  point  of  the  car-frame)  is  passing  through 
the  distance  s  (parallel  to  rail),  the  work  done  on  the  rail-pressure, 

R,  is  not  zero,  but  =  R .  —s.     It  is  convenient,  therefore,  in 


or 


NOTES   AND   EXAMPLES   IN   MECHANICS. 

applying  the  principle  of  work  and  energy  to  the  case  of  a  car, 
to  ^nsider  that-  the  rolling  is  perfect  and  that  the  work  actually 
due  to  rolling  resistance  is  occasioned  by  the  overcoming  of  an 

imaginary  backward  force,  -7?,  applied  directly  to  the  car-frame, 
centre  of  axle  of  wheel  ;  since  the  product     —  E  #  is  equal  to 

that  of  R  by  f-sj.     This  imaginary  force  is  what  is  referred  to 

in  the  foot-note  of  p.  189,  M.  of  E. 

95.  Examples  of  §  173,  M.  of  E.  —  In  Example  2,  since  slipping 
is  impending,  the  backward  tangential  action,  or  force,  of  the  rail 
on  each  wheel  must  be  T,  —  (0.20  x  5000)  Ibs.  ;  and  since  its 
moment,  Tr,  about  the  wheel-axis  must  balance  that,  J?rt  of  the 
friction  ^between  brake  and  wheel-rim  (neglecting  the  moment 
of  the  small  axle-friction),  this  friction  must  be  F  —  1000  Ibs. 

When  the  car-frame  moves  through  the  unknown  distance  s 
before  coming  to  rest,  the  friction  F  (an  internal  friction)  is  over- 
come through  a  relative  distance  measured  on  the  wheel-tread  of 
length  =  s  (because  the  wheels  do  not  slip  and  the  brake-shoe 
presses  on  the  same  rim  as  that  which  rolls  on  the  rail  ;  whereas, 
if  the  shoe  rubbed  on  the  rim  of  a  drum  on  the  same  axle,  the 
circumference  of  the  drum  being  (for  example)  one  half  that  of 
the  wheel-rim,  the  relative  distance  rubbed  through  would  only 
be  one  half  of  s. 

Gravity  being  neutral  and  both  axle  and  rolling  resistances 
being  neglected,  we  have,  from  eq.  (XVI),  p.  149,  M.  of  E.,  (with 
foot,  pound,  and  second,) 


0  -  8  X  1000  X  s  =  0  -  (80)';     /.  s  =  496  ft 

&  oz.z 

In  Example  3,  although  the  track  is  on  an  up-grade,  or 
inclination  <*,  with  the  horizontal,  the  angle  is  so  small  that 
60  -T-  5280  may  be  taken  as  either  its  nat.  tang,  or  its  sine,  at 
will.  The  weight  40,000  Ibs.  is  here  a  resistance,  being  raised 


EXAMPLES   IN    DYNAMICS.  103 

through  a  height  of  1000  ft.  X  sin  a,  =  11.35  ft.,  and  both  roll- 
ing and  axle  resistances  are  neglected.     Hence 

0  _  SF  X  1000  ft.  -  40000  Ibs.  X  11.35  ft.  =  0  -     •  - 


or      =         Ibs. 

Example  4  differs  from  the  preceding  only  in  the  introduc- 
tion of  two  more  items  of  negative  work.  The  work  of  rolling 
resistance  is  ascribed  to  the  overcoming  of  an  imaginary  force 

—  I-^K  X  5000]  Ibs.,  =  6.66  Ibs.  for  each  wheel  (applied  to  car- 
\  15  / 

frame),  through  a  distance  of  1000  feet. 

The  axle-friction  at  the  journal  of  each  wheel  is  (0.036  X  5000) 
Ibs.  =  180  Ibs.  (taking  the  coefficient  near  the  top  of  p.  190, 
M.  of  E.).  A  point  in  the  circumference  of  the  journal  rubs 

through  a  distance  of  LjA  X  1000  ft.  =  100  ft.     Hence,  finally, 

—  8^x1000  —  40000  X  11.35-8x6.66  X  1000—8x180  X  100 


96.  Examples  in  Dynamics.     (Statements  on  p.  194,  M.  of  E.) 
—  Example  1.     Fig.  117  shows  the  end  of  the  shaft  in  question. 
The  friction   on  the  side,  or  u  axle  friction,"  is 
to   the    pressure   of   40    tons,    =  N'  ;    i.e.,  ( 
fN'  =  .05  X  80,000  =  4000  Ibs.,  and   is  \ 


due  to  the  pressure  of  40  tons,  =  N'  ;  i.e.,  ,'  •  *~-  N 
f"  =fN'  =  .05  X  80,000  =  4000  Ibs.,  and  is  \  * 
overcome  through  a  circumference  of  TT  x  1  ft.,  N'||  |  J40*on«v 
—  3.14  ft.,  each  revolution.  The  work  done  per  *'io.  nr. 
revolution  on  the  friction  at  the  base  (due  to  the  10  tons)  is  the 
same  as  if  all  concentrated  at  a  distance  =  f  of  the  radius 
from  the  centre,  and  is  therefore  (.05  X  20000)  X  (|TT  X  1)  — 
(1000  Ibs.)  X  (2.094  ft.).  Hence  the  lost  work  per  second,  i.e., 
the  power,  absorbed  by  friction,  in  ft.  -Ibs.  per  sec.,  is 

f£[4000  X  3.14  +  1000  X  2.094]  =  12211  =  22.2  H.  P, 


104 


NOTES   AND   EXAMPLES   IN   MECHANICS. 


FIG.  118. 


Example  2.  Fig.  118.  (This  form  of  friction-brake  is  some- 
times used  in  testing  motors.)  Since  slipping  ac- 
tually occurs,  we  use  eq.  (3)  of  p.  183,  M.  of  E., 

to  — i 
-£   .     Here   we   put 

Sn  =  20  Ibs.,  /&„  —  10  Ibs.,  with  a  =  TT,  and  obtain 

f=  [2.302  X  Iog10(2.0)]  -  ft,  =  0.22. 
As  to  the  power,  Z,  expended  on  the  friction,  we 
note  that  each  element  of  the  friction  is  overcome  through  a  dis- 
tance v  every  second,  where  v  is  the  linear  velocity  of  the  pulley- 
rim;  i.e.,  L  =  Fv\  denoting  the  sum  of  the  elementary  frictions 
by  F.  These  elementary  frictions  are  not  parallel,  but  we  note 
that  each  (being  tangential)  has  the  same  lever-arm,  =  9  inches, 
—  r,  about  the  centre  of  the  circular  curve,  so  that  when  the 
curved  part  of  the  strap  is  considered  free,  and  moments  are 
taken  about  that  centre,  we  obtain  Fr  —  Snr  —  S0r  (the  normal 
pressures  on  the  cord  have  zero  moments) ;  hence  F  =  Sn  —  S9 
and  the  power  in  question  =  L  =  [/Sn  —  S9~]v.  Therefore 

L  =  [(20  -  10)  Ibs.]  X  [(27T  x  TV  X  Y<J°-)  ft.  per  sec.]  =  78.5 
ft-lbs.  per  sec. ;  =  0.142  H.  P. 

Should  the  pulley  be  made  to  turn  the  other  way  with  the 
same  speed,  the  weight  of  20  Ibs.  (now  =  /#„)  remaining  the  same, 
more  power  will  be  required.  Assuming  the  coefficient  of  fric- 
tion between  strap  and  pulley  to  remain  unchanged,  the  spring- 
balance  will  now  read  40  Ibs.  tension  (—  8n),  since  ef*  =  2,  so 
that  Sn  —  So=  20  Ibs.  instead  of  10  as  before,  v  being  the  same 
(7.85  ft.  per  sec.),  the  power  will  be  double  that  previously 
found ;  viz.,  L  now  =  0.282  H.  P. 

Example  3.    Fig.  119.     It  is  assumed  that  the  pressure  on  the 
journals,  or  axle,  is  due  solely  to  the  weight,  6?, 
of  the  stone ;  i.e.,  that  the  caps  of  the  bearings  are 
not  clamped  in  contact  with  the  journals ;  other- 
wise the  friction  would  be  >/'#,  for  axle-friction. 

The  stone  being  considered  free,  we  note  that 
its  weight  and  the  normal  component  of  the  reac- 


FIG.  119. 


tion  of  the  bearing  are  neutral  forces,  the  friction,  or  tangential 


EXAMPLES 


DYNAMICS. 


105 


component,/'  #,  is  a  resistance  ;  and  that  there  are  no  working 
forces. 


The  initial  K.  E.*  of  the  stone  is 


>*—&?  (o?0  being  the  initial 

i7 


angular  velocity)  ;  its  final  K.  E.,  zero.  Hence  the  initial  K.  E. 
is  all  absorbed  in  the  work  of  friction.  Since  the  number  of 
turns  in  coming  to  rest  is  160,  the  total  distance  through  which 
the  friction  at  the  circumference  of  the  journal  is  overcome  is 
160  X  7r[1.5  -r-  12]  =  62.85  ft.  "With  the  foot  and  second,  GOQ  = 
27T  x  -W0-  =  4?r  radians,  and  g  =  32.2.  Therefore,  from  the 
Principle  of  Work  and  Kinetic  Energy  [eq.  (XV),  p.  147,  M. 
of  E.], 


P-? 


Example  4.     To  move  A  horizontally  with  an  acceleration 
=  15  (foot  and  sec.)  would  require  a  horizontal  force  =  mass  X 

2 
ace.  =  oo~o  X  15  =  .93  Ibs.     But  the  horiz.  compon.  (friction) 

of  the  action  of  E  on  A  is  only  —  fN  —  0.3  X  2  =  0.6  Ibs., 
which  is  <  .93.  Hence  A  will  riot  keep  abreast  of  B,  but  will 
gradually  fall  behind  B. 

97.  Brake-strap,  Lever,  and  Descending  Weight,  Numerical 
Example  (Fig.  120).—  The  weight  Q  of  600  Ibs.  is  to  be  let  down 
without  acceleration,  the  rope  un- 
winding from  a  drum  of  1  ft. 
radius.  On  the  shaft  of  the  drum 
is  keyed  a  pulley  of  2  ft.  radius, 
the  friction  on  whose  rim,  due  to 
its  rubbing  under  the  encircling 
stationary  strap,  can  be  varied  in 
amount  by  a  force  P  exerted  on 
the  lever  AB.  It  is  required  to 
compute  a  proper  value  for  the 
pressure  P  in  the  present  instance 
to  prevent  acceleration  of  the  600-lb.  weight,  the  coefficient  of 
friction  of  the  strap  on  the  pulley  being  assumed  to  bef  =  0.30. 

*  Kinetic  Energy. 


FIG.  120. 


106  NOTES   AND   EXAMPLES   IN   MECHANICS. 

The  strap  covers  three  quarters  of  the  pulley-rim  (i.e.,  a  =  f  TT). 
See  figure  for  other  dimensions. 

If  Q  sinks  without  acceleration,  the  tension  in  the  vertical 
part  of  the  rope  must  be  600  Ibs.,  and  the  rotation  of  pulley  be 
uniform  ;  hence  moments  must  balance  for  the  pulley,  drum,  and 
shaft  ;  so  that  (with  the  foot,  pound,  and  second)  600  X  V  =< 
F  X  2',  where  ./'"is  the  sum  of  the  requisite  frictions  (tangential 
forces)  of  strap  on  pulley  ;  i.e.,  F  =  300  Ibs. 

But  from  the  equilibrium  of  the  curved  portion  of  strap,  by 
moments  about  centre  of  the  curve,  S0x  ^'  —  SnX  2'-|-^X  2'=  0. 
8n  is  the  tension  in  the  vertical  straight  part  of  the  strap;  /$„, 
that  in  the  horizontal.  Sn  is  greater  than  S0  and  bears  to  it  the 

/  o  \ 

relation  Sn  —  S0ef«  ;  or,  \fit  =  loge  (~J.     That  is  v^e  p.  184, 

M.  of  E.),  (Sn  :  S0)  =  the  number  whose  common  logarithm  is 
(.45)*  X  0.43429  =  4.12.  Combining  this  with  F  =  Sn  —  S0 
(see  above),  we  have  (4.12  —  l)S0  =  F  —  300  Ibs.;  whence 
£o  =  300  -j-  3.12  =  96.15  Ibs.  ;  and  ^n,  =  4.12^0,  =  396.13  Ibs. 
The  requisite  force  P  is  then  found  by  noting  that  for  the 
equilibrium  of  the  lever,  the  moments  of  the  three  forces  Sn,  £0, 
and  P,  must  balance  about  the  fulcrum  A  ;  i.e., 


PX^^^xi  +  ^Xi;     or     P  =  £(&  +  #.)  =  61.5  Ibs 
a  pressure  easily  applied  by  one  man. 


CHAPTER  VII. 
MECHANICS  OF  MATERIALS  AND  GRAPHICAL  STATICS. 

98.  Intensity  of  the  Shearing,  and  of  the  Normal,  Stress  on  an 
Oblicme  Section  of  a  Prism  under  Tension.  (This  treatment  may 
be  more  readily  understood  than  that  now  given  in  §  182  of  M. 
of  E.) — From  the  prism  under  tension  in  Fig.  193,  M.  of  E.,  con- 
sider by  itself  a  portion  shown  in 
Fig.  121,  between  the  right  section 
J£N  &\\&  any  oblique  section,  ML. 
The  area  of  the  right  section  being 
F  and  the  intensity  stress  per  unit 
area  of  that  section  being  p,  Fp 
expresses  the  total  stress  on  the  FlG- 

right  section.  The  total  stress  on  ML  is  also  of  course  =  Fp. 
Its  component  P"  normal  to  the  plane  ML  is  evidently  P"  = 
Fp  sin  of.  This  is  the  total  normal  stress  on  this  oblique  plane 
ML.  But  the  area  ML  over  which  this  normal  stress  is  dis- 
tributed is  not  =  F,  that  of  the  right  section  of  the  prism,  but 
=  (F  -T-  sin  a).  Hence  to  obtain  the  normal  stress  on  ML  per 
unit  of  area,  i.e.,  the  intensity  of  normal  stress  on  ML^  we  must 
divide  P"  by  (F  ~  sin  a)  and  thus  obtain  : 

Normal  stress,  per  unit  of  \ 

•  c  — •  T)   sin    OL  (I  i 

area,  on  oblique  section  \       ^ "  *  ' 

Similarly,  the  other  component,  P'  (of  the  total  stress  on 
ML),  which  is  tangential  to  ML,  is  in  amount  —  Fp  cos  a,  which 
is  the  total  shearing  stress  on  ML.  To  obtain  the  intensity  of 

107 


108  NOTES   AND   EXAMPLES   IN   MECHANICS. 

this  shearing  stress,  we  divide  Pf  by  the  area  (F r-s-  sin  «)  of  ML, 
over  which  P'  is  distributed,  and  obtain : 

Shearing  stress,  per  unit  \ 

'^  \  =  p.  sin  a  cos  a.       .     (2V 

of  area,  on  oblique  section  ) 

In  these  two  equations  p  is  an  abbreviation  for  P  -~-  F,  and 
a  is  the  angle  that  the  oblique  plane  ML  makes  with  the  axis  of 
the  prism. 

The  reason  for  ascertaining  the  stress  per  unit  area  in  any 
case  is,  of  course,  that  the  safety  of  the  material  depends  upon  it 
and  not  simply  on  the  total  stress. 

99.  Spacing  of  Rivets  in  a  Built  Beam. — The  statement  in  the 
middle  of  p.  293  of  M.  of  E.  that  "  The  riveting  connecting  the 
angles  with  the  flanges  (or  the  web  with  the  angles),  in  any 
locality  of  a  built  beam,  must  safely  sustain  a  shear  equal  to  J 
(the  total  shear  of  the  section)  on  a  horizontal  length  equal  to  the 
height  of  the  web"  may  be  most  directly  utilized  as  follows : 
Imagine  the  horizontal  continuity  of  the  web  to  be  broken  and 

consequently  a  vertical  seam  ren- 
dered necessary,  as  shown  in  Fig. 
122. 

Then,  whatever  spacing  of  rivets 
would  be  necessary  in  this  ideal  seam 
can  be  adopted  in  the  real  horizontal 
FlG- 182-  seam  made  by  riveting  together  the 

web  and  angles,  the  rivets  being  considered  to  be  in  double  shear 
(or  single)  in  the  ideal  case,  if  so  in  the  actual.  For  example, 
taking  the  data  of  the  example  on  p.  294,  M.  of  E.,  there  must 
be  enough  rivets  in  the  vertical  seam,  of  length  =  h0  =  the 
height  of  web,  to  safely  stand  the  total  shear  of  J  =  40,000  Ibs. 
Since  each  rivet  can  safely  endure  a  shear  of  9000  Ibs.  in  double 
shear  (see  p.  294),  the  number  of  rivets  required  in  the  height  of 
web  would  be  40000  -^-  9000  =  4.44 ;  i.e.,  they  should  be  spaced 
4.5  in.  apart  since  the  height  of  web  is  20  in.,  and  20"  -f-  4.44  = 
4.5". 

But  since  the  pressure  on  the  side  of  each  hole  is  limited  to 
i>470  Ibs.  (see  p.  294),  on  this  basis  the  number  of  rivets  in  height 


MECHANICS   OF   MATERIALS.  109 

of  web  should  be  40,000  -f-  5470  =  7.2,  which  implies  putting 
them  at  a  distance  apart,  centre  to  centre,  of  20  -r-  Y.2  =  2.7  in. 
This  spacing,  therefore,  should  be  adopted  in  the  horizontal  seam 
between  web  and  angles  at  this  part  of  the  beam  (near  abutment). 
100.  I-beams  treated  without  the  Use  of  the  Moment  of  Inertia, 
— The  assumption  is  so  frequently 
made  (for  simplicity  in  treating  the 
web)  that  the  web  carries  all  the 
shear,  that  the  corresponding  as- 
sumption  that  the  two  flanges  carry 
all  the  tension  and  compression  is 
also  sometimes  used  to  simplify  the 
analysis.  With  thin  webs  the  re- 
sults obtained  are  accurate  enough 
for  practical  purposes.  For  ex- 
ample, suppose  a  horizontal  I-beam  to  rest  upon  supports  at  its 
extremities  and  to  bear  several  detached  loads.  To  find  the 
tensile  or  cornpressive  stress  in  the  flanges  at  any  right  section  as 
m,  consider  free  the  part  of  beam  on  the  left  of  that  section  (see 
Fig.  123).  R  is  the  abutment  reaction ;  Pl  and  P9  the  loads 
between  m  and  A.  Considering  the  total  compression  P'  to  be 
uniformly  distributed  over  the  area  of  the  upper  flange,  and 
hence  (for  geometrical  purposes)  to  be  applied  in  the  centre  of 
gravity  of  that  flange,  and  similarly  the  total  tension  P"  in  the 
centre  of  gravity  of  the  lower  flange  (h!  •=.  vertical  distance  be- 
tween), while  the  web  carries  the  shear  alone ;  we  obtain  (by 
taking  moments  about  the  intersection  of  <7and  P"\ 


from  which  P'  is  found.  Evidently  P"=  P ',  since  they  are  the 
only  forces  having  horizontal  components.  Let  F'  =  the  area  of 
either  flange,  then  the  stress  per  unit  area  in  either  flange  is 

P' 
jp'  =  -p-,.     If  this  result  is  too  large  for  safety  the  flange-area 

must  be  increased. 

By  this  method,  the  proper  amount  of  sectional  area  can  be 
computed  for  the  flanges  at  each  of  several  sections  of  a  beam  to 


110  NOTES   AND   EXAMPLES   IN   MECHANICS. 

carry  a  specified  loading,  and  a  beam  of  "  uniform  strength"  be 
designed.  In  such  a  case,  however,  the  weight  of  the  beam  itself 
cannot  be  estimated  in  advance,  but  after  a  gradual  tentative 
adjustment  can  be  taken  into  account  in  a  satisfactory  manner. 

(The  upper  flange  being  in  compression  may  need  to  be 
braced  or  latticed  with  those  of  the  accompanying  parallel  beams 
to  prevent  horizontal  buckling.)  Plate  girders  of  variable  section 
ape  designed  on  this  general  principle,  the  flange-area  being 
increased  toward  the  middle  of  the  span  by  riveting  on  additional 
plates. 

101.  Load  of  "  Incipient  Flexure"  of  Long  Slender  Columns,— 
The  result  brought  out  in  eq.  (8)  of  p.  366,  M.  of  E.,  that  in  the 
case  of  a  long,  slender,  round-ended,  prismatic  column,  no  flexure 
at  all  takes  place  until  a  definite  value  for  the  load  is  reached,  and 
that,  with  that  value,  any  small  deflection  whatever  can  be  main- 
tained, may  be  arrived  at  quite  rationally  as  follows,  without 
intricate  analysis  : 

Let  the  horizontal  bed-plates  of  a  testing  machine,  Fig.  124, 
be  advanced  toward  each  other  until  the  slender  round- 

m 

ended  rod  AB  between  them  is  deflected  a  small  amount 
=  a  at  the  middle.  If  p  is  the  radius  of  curvature  of  the 
elastic  curve  at  R,  we  have  (from  free  body  7?  .  .  A) 


T>  V 

-  —  Pa  ;     or    P  =  —  . 
p  ap 

(This  radius  of  curvature,  p,  is  nearly  =  to  that  of  the 
circle  determined  by  the  three  points  B,  It,  and  A  (T%- 
of  it). 

Now  let  the  plates  be  separated  until  the  deflection  is 
only  =  a'  ;  call  the  new  radius  of  curvature,  p'.    Then  the 

EI 

FIG.  124.  new  value  of  the  force  or  pressure  at  each  end  is  P'  =  -7—,. 

CL  fJ 

But,  for  this  very  flat  curve,  as  the  deflection  is  diminished,  the 
radius  of  curvature  changes  in  an  inverse  ratio  ;*  i.e.,  p'  :  p  :  :  a  :  a'  ; 
or,  ap  =  a'  p'.  Therefore  P  =  P'  ;  i.e.,  the  pressure  induced  ly 
the  elasticity  of  the  rod  against  the  plates  does  not  diminish  with 
a  diminishing  deflection,  hut  remains  constant.  Or,  conversely, 
if  a  less  force  than  this  is  applied  to  the  column,  no  deflection 

*  Strictly,  the  relation  is 

a(2p  —  a)  =  [^Jj?]3,  =  practically  a  constant. 


TESTS   OF   WOODEN   POSTS.  Ill 

takes  place  ;  while  if  an  actual  load,  whose  weight  is  greater  than 
the  above  force,  be  placed  on  the  column,  the  upward  pressure 
against  it  due  to  the  elasticity  of  the  column  (as  the  latter  bends) 
being  always  less  than  the  weight  (unless  perhaps  when  the  de- 
flection becomes  extreme),  the  load  sinks  with  an  accelerated 
motion  and  the  column  finally  breaks  (since  with  increased  deflec- 
tion the  stress  in  the  outer  fibre  is  augmented). 

In  most  cases  in  practice,  columns  are  not  sufficiently  slender 
to  bear  out  all  the  above-described  phenomena,  but  enough  has 
been  said  to  show  that  while  with  a  horizontal  beam  the  deflec- 
tion is  nearly  proportional  to  the  load  applied  (within  elastic 
limit),  such  is  far  from  being  the  case  with  a  column. 

If  a  piece  of  card-board  cut  from  a  visiting  card,  and  about 
half  an  inch  wide  by  three  or  four  inches  long,  be  pressed  end- 
wise with  the  finger  on  the  scale-pan  of  a  letter-scale,  the  value  of 
the  pressure  corresponding  to  different  deflections  can  be  easily 
noted  and  the  above  claims  roughly  verified. 

102.  Recent  Tests  of  Large  Wooden  Posts — Prof.  Lanza,  of 
the  Massachusetts  Institute  of  Technology,  Boston,  has  made 
tests  on  yellow-pine  flat-ended  posts,  giving  results  as  follows : 
Highest  breaking  stress  5400  Ibs.  per  sq.  in. ;  lowest,  3600 ; 
average,  4544. 

These  yellow-pine  posts  were  nearly  cylindrical  in  form  and 
almost  all  of  them  12  ft.  in  length  (a  few  2  ft.  long);  diameters, 
from  6  to  10  in. 

With  white  oak  posts,  flat-ended,  and  of  about  the  same  sizes 
as  the  former,  the  highest  breaking  stress  was  4600,  the  lowest, 
3000  Ibs.  per  sq.  in. ;  with  exception  of  two  which  reached  6000. 
In  all  these  cases  of  pine  and  oak  posts  failure  occurred  by  direct 
crushing,  lateral  deflections  being  inconsiderable,  showing  that 
all  were  practically  " short  Hocks" 

Eight  separate  tests  were  made  with  the  load  applied  eccen- 
trically, about  two  inches  off  the  centre,  the  result  being  to 
diminish  the  strength  by  about  one  third.  All  these  posts  had 
been  in  use  for  years  and  were  well  seasoned.  Each  had  a  hole 
about  2  in.  in  diam.  along  the  axis,  from  end  to  end. 

Other  tests  have  been  made  at  Watertown,  Mass.,  with  the 


112  NOTES   AND   EXAMPLES   IN   MECHANICS. 

Government  testing  machine  on  timber  columns,  of  rectangular 
sections,  mostly  about  5  by  5  in.,  and  7.5  by  9.7  in. ;  with  a  num- 
ber 5  by  15  in.,  and  8  by  16  in.  Their  lengths  ranged  progress- 
ively from  15  in.  to  27  ft.  Flat-ended  supports.  From  these 
tests  Mr.  Ely  concludes  that  if  the  breaking  load  in  pounds  be 
expressed  as  P  =  FC,  where  F  is  the  sectional  area  in  square 
inches,  one  of  the  following  values  for  C  should  be  taken  accord-' 
ing  to  the  kind  of  timber  and  the  ratio  of  the  length  I  to  the 
least  side,  £>,  of  the  rectangular  section  ;  thus  : 

For  white  pine  : 

For  I  H-  b  =  0  to  10 ;  10  to  35  ;  35  to  45  ;  45  to  60, 
Make      C=   2500;        2000;         1500;         1000. 

For  yellow  pine : 

For  Z  -^-  5  =  0  to  15;  15  to  30;  30  to  40;  40  to  45;  45  to  50;  50  to  60, 
Make  G=  4000;  3500;  3000;  2500;  2000;  1500. 

103.  The  Pencoyd  Tests  of  Full-size  Rolled-iron  I-beams, 
Channels,  Angles,  Tees,  etc.,  used  as  Columns. — These  were  made 
in  1883  by  Mr.  Christie,  at  the  Pencoyd  Iron  Works  of  Phila- 
delphia, Pa.,  and  were  very  careful  and  extensive.  The  following 
table  is  based  on  them.  By  "fixed  ends"  it  is  here  meant  that 
the  ends  are  so  securely  attached  to  the  contiguous  supports  that 
the  fastenings  would  not  be  ruptured  if  the  column  were  sub- 
jected to  a  breaking  load;  by  "flat  ends,"  that  the  ends  are 
squared  off  and  bear  on  a  fiat,  firm  surface.  "  Hinged  ends " 
refers  to  the  ends  being  fitted  with  pins,  or  ball-and-socket  joints, 
of  proper  size,  with  centres  practically  in  the  axis  of  the  column 
(this  axis  being  the  line  containing  the  centres  of  gravity  of  all 
sections  of  the  column,  which  is  prismatic);  while  "round  ends" 
implies  that  the  ends  have  only  points  of  contact  such  as  balls  or 
pins  bearing  on  a  flat  plate,  the  point  of  contact  being  in  the 
axis  of  the  column.  The  first  column  of  the  table  contains  the 
ratio  of  the  length,  £,  of  the  column  to  #,  the  least  radius  of  gyra- 
tion of  the  section  (except  that  for  hinged  ends  the  pin  must  be 
at  right  angles  to  the  least  radius  of  gyration).  Factors  of  safety 
are  recommended  as  given,  being  different  for  fixed  and  flat  ends 


THE    PENCOYD    EXPERIMENTS    WITH   COLUMNS. 


113 


Ratio 
I 

k 

Fixed  Ends. 

Flat  Ends. 

Factor  of 
Safety. 

Hinged  Ends. 

Round  Ends. 

Factor  of 
Safety. 

20 

46,000 

46,000 

3.2 

46,000 

44,000 

3.3 

40 

40,000 

40,000 

3.4 

40,000 

36,500 

3.6 

60 

36,000 

36,000 

3.6 

36,000 

30,500 

3.9 

80 

32,000 

32,000 

3.8 

31,500 

25,000 

4.2 

100 

30,000 

29,800 

4.0 

28,000 

20,500 

4.5 

120 

28,000 

26,300 

4.2 

24,300 

16,500 

4.8 

140 

25,500 

23,500 

4.4 

21,000 

12,800 

5.1 

160 

23,000 

20,000 

4.6 

16,500 

9,500 

5.4 

180 

20,000 

16,800 

4.8 

12,800 

7,500 

5.7 

200 

17,500 

14,500 

5.0 

10.800 

6,000 

6.0 

220 

15,000 

12,700 

5.2 

8,800 

5,000 

6.3 

240 

13,000 

11,200 

5.4 

7,500 

4,300 

6.6 

260 

11,000 

9,800 

5.6 

6,500 

3,800 

6.9 

280 

10,000 

8,500 

5.8 

5,700 

3,200 

7.2 

300 

9,000 

7,200 

6.0 

5,000 

2,800 

7.5 

320 

8,000 

6,000 

6.2 

4,500 

2,500 

7.8 

340 

7,000 

5,100 

6.4 

4,000 

2,100 

8.1 

360 

6,500 

4,300 

6.6 

3,500 

1,900 

8.4 

380 

5,800 

3,500 

6.8 

3,000 

1,700 

8.7 

400 

5,200 

3,000 

7.0 

2,500 

1,500 

9.0 

420 

4,800 

2,500 

2,300 

1,300 

440 

4,300 

2,200 

2,100 

460 

3,800 

2,000 

1,900 

480 

1,900 

1,800 

from  those  proposed  for  hinged  and  round  ends.  In  the  other 
columns  the  numbers  given  are  the  respective  breaking  stresses 
in  Ibs.  per  sq.  in.  of  sectional  area,  which  number  must  be  multi- 
plied by  that  area  in  sq.  in.,  for  the  actual  breaking  load  ;  divid- 
ing which  by  the  proper  factor  of  safety  we  obtain  the  safe  load 
in  pounds. 

For  example,  required  the  breaking  load  of  a  9-in.  light  iron 
I-beam  of  the  N.  J.  Steel  and  Iron  Co.,  14  ft.  long  and  used  as  a 
column  with  flat  ends. 

From  p.  40  of  the  hand-book  we  find :  least  /  =  4.92  and 
the  area  of  section  =  7  sq.  in.  Hence  the  least  k2  is  I  ~  F, 
=  0.703,  and  k  itself  =  0.838  in. ;  so  that  I  -+-  k  =  168  in.  -f- 
0.838,  =  200.  The  breaking  load,  then,  is  FC  =  7.00  X  14,500 
=  101,500  Ibs. ;  and  the  safe  load  would  be  101,500  -r-  5  —  20,300 
Ibs.  (If  Rankine's  formula  were  applied  to  this  same  case  a  fair 
agreement  with  this  result  would  be  found.) 


114 


NOTES   AND    EXAMPLES   IN   MECHANICS. 


104.  Cast-iron   Channel  as  a  Column. — In    the   case   of   the 
,^  channel-shaped  section  in  Fig.  125,  com- 

posed of  three  rectangles,  all  of  the  same 
width,  —  t,  it  is  desirable,  for  economy  of 
material  if  the  prismatic  body  is  to  be 
used  as  a  column,  that  the  moments  of 
inertia  about  the  two  gravity  axes  X  and 
T  should  be  equal  (see  §  311,  M.  of  E.). 
The  problem,  therefore,  presents  itself  in 
this  form  :  Given  the  width  5  of  the  base 
DC  of  the  section,  and  the  thickness  of 
metal  t,  what  value  should  be  given  to  the 
length  c,  of  the  projecting  sides  AD  and 
BG,  of  the  channel  to  secure  this  result? 

The  algebraic  statement  of  the  conditions  involved  leads  to 
an  equation  of  high  degree  for  the  unknown  quantity  c.  But, 
~by  numerical  trial,  a  few  reliable  values  have  been  found,  given 
below,  by  simple  interpolation  between  which  all  ordinary  cases 
in  practice  may  be  satisfied. 


X 

1 

. 

, 
c 

•  >: 

"f 

| 
I 

I 

i 

! 

i 

1 

SL£ 

I 

D 

—  -> 

[— 

~> 

c 

FIG.  125. 


When  t  =  0  (infinitely  thin), 
"      t  —  0.0835  —  ^5, 
"      t  =  0.1665  =  £5, 

j      solid 
I  rectangle 


make  c  =  1.375 ; 
«      0  =  1.305; 


t  —  0.5005  = 


r«     G  =  1.005. 


In  the  construction  of  such  a  column  the  edges  A  and  B  of 
the  projecting  sides  should  be  tied  or  braced  together  at  inter- 
vals ;  or  occasional  transverse  webs  may  be  introduced. 

It  is  remarkable  in  this  problem  that  the  distance  u  of  the 
centre  of  gravity  C  from  the  base  DC  is  almost  exactly  equal  to 
one  half  of  5  in  every  instance,  and  may  be  so  assumed  in  locat- 
ing the  axis  of  the  column  that  the  load  may  be  applied  in  that 
axis. 

105.  Vertical  Reactions  of  Horizontal-faced  Piers  bearing  a 
Beam  and  Loads.  Graphical  Method. — The  construction  on  p.  404 
of  M.  of  E.  is  the  one  usually  given  and  is  the  most  convenient; 
still,  the  proof  is  a  little  puzzling  to  the  student  because  the 


PIER- REACTIONS — GRAPHICS. 


115 


amounts  and  direction  of  the  two  auxiliary,  mutually  annulling, 
forces  P  and  P'  are  not  known  at  the  outset. 

Hence  we  here  present  a  construction  in  which  those  two 
forces  are  completely  assumed  and  known  at  the  outset.  Assume 
1  and  6  as  the  auxiliary  forces  (equal,  opposite,  and  in  the  same 
line),  there  being  three  loads  in  this  instance,  2,  3,  and  4;  Fig. 
126,  Number  the  forces  of  the  system  as  in  figure  and  draw  a 


FIG.  126. 

portion  of  the  force-polygon  with  the  known  forces  1,  2,  3,  and 
4,  beginning  at  O,  and  the  first  three  rays,  I,  II,  and  III  (dotted). 
The  first  segment  of  the  corresponding  equilibrium  polygon 
should  begin  at  #,  the  intersection  of  the  action-lines  of  forces  1 
and  2,  and  finally  the  third  segment  cuts  the  action-line  of  5  in 
the  point  m.  Now  the  fourth  segment  is  the  last  in  this  case  [of 
three  loads,  2,  3,  4],  and  must  pass  through  the  intersection  of 
the  last  two  forces,  6  and  7,  i.e.,  through  A.  Hence  draw  mA 
and  a  parallel  to  it  through  the  pole  O,  this  latter  line  being  the 
fourth  ray  desired,  whose  intersection  with  the  "  load-line"  at  nf 
cuts  off  the  proper  length  of  the  right-hand  reaction  5.  The 
forces  6  and  7  are  now  easily  added  to  the  force-diagram  in  an 
obvious  manner  and  the  latter  is  complete  ;  the  magnitude  of  the 
other  reaction,  7,  being  thus  determined.  Of  course,  the  value 
of  force  7  is  also  given  by  the  rid,  and  the  force-polygon  could 
also  be  closed  by  running  from  nf  to  d  and  then  from  d  to  0. 
instead  of  in  the  manner  shown. 

Note  the  order  of  numbering  in  the  above.  The  two  assumed 
forces  are  made  the  first  and  the  last  but  one,  respectively ;  the 
unknown  reactions  are  made  the  last  ~but  two,  and  the  last, 
respectively ;  while  the  given  loads  are  assigned  to  the  interven- 
ing numbers,  in  any  order. 


116 


NOTES   AND   EXAMPLES   IN   MECHANICS. 


(in.) 


106.  Construction  for  Use  with  the  Treatment  in  §  390,  M.  of 
E.  (Graphical  Statics). — For  the  particular  case  involved  in  the 
treatment  of  the  straight  horizontal  girder,  built  in,  of  §  390, 
M.  of  E.,  the  following  is  given,  to  replace  the  more  general 
construction  of  §  377,  M.  of  E. 

At  (I.)  in  Fig.  127,  we  have  a  curve,  or  broken  line  (equi- 

„  librium  polygon,  for  instance), 
-1>v*./  I  FKWMG,  connecting  points  F 
\^>^  m  and  G  in  two  vertical  lines. 
Across  this  curve  we  wish  to  draw 
a  right  line  v  .  .  m,  in  such  a  way 
that  the  area K WM  above  v  .  .m 
shall  be  equal  to  the  sum  of  the 
areas  vFK  and  mGM  below 
v  .  .  m,  and  also  that  the  centre 
of  gravity  of  the  upper  area  shall 
be  in  the  same  vertical  line  as 
that  of  the  two  lower,  or  negative 
areas,  taken  together. 

Only  one  position  of  v  .  .  m 
will  do  this,  the  algebraic  expression  for  which  is  that  2(s*)  =  0, 
and  that  2(%z')  =  0 ;  (the  areas  in  question  being  divided  into 
vertical  strips  of  equal  horizontal  widths  =  4x9  the  distance  of 
any  strip  from  the  vertical  line  vF  being  called  a?.) 

Annex  the  figure  FGMK  (having  joined  F .  .  G)  to  both 
the  positive  and  negative  figures  above  mentioned  and  the  con- 
dition now  becomes  [see  (II.)  in  figure]  that  the  area  of  the 
figure  FKWMG..F  (the  right  line  FG  being  its  lower 
boundary)  must  equal  that  of  the  two  triangles  vFG  and  mGv, 
and  that  the  centre  of  gravity  of  the  former  must  lie  in  the  same 
vertical  as  that  of  the  two  triangles  combined.  In  other  words, 
if  the  area  of  the  curvilinear  figure,  FKWMG-F,  with  FG  as 
base,  be  considered  as  a  weight  R  acting  through  its  centre  of 
gravity,  then  the  areas  of  the  two  triangles  must  represent  the 
two  upward  reactions  T&ud  T'  of  two  piers  [see  (III.)  in  figure] 
supporting  a  horizontal  beam  on  which  R  rests.  These  imagi- 
nary piers  are  evidently  at  distances  of  one  third  the  span  from 


| 


FIG.  127. 


EQUILIBRIUM  POLYGON  THROUGH  THREE  POINTS.     117 

the  verticals  through  F  and  G.  Adopt,  therefore,  the  following 
construction : 

By  dividing  into  vertical  strips  find  the  area  of  the  curvi- 
linear figure  with  base  FG.  Draw  the  pier  verticals  at  the  one- 
third  points.  Find  the  vertical  containing  the  centre  of  gravity 
of  the  curvilinear  figure  by  p.  415.  Compute  or  construct  the 
values  of  T  and  T'  on  the  conception  of  the  known  area  R  being 
a  weight  supported  on  the  beam  with  T  and  T'  as  reactions. 
(T  and  T'  are  most  easily  obtained,  perhaps,  by  scaling  the 
distances  s  and  t  (see  figure)  and  writing  T(s  +  t)  =  Rt,  and 
T'  =  R  -  T.) 

If  the  area  of  the  triangle  FGv  must  be  =  T,  one  of  the 
values  just  found,  knowing  that  this  area  =  \  (altitude  1}  X  base 
vF,  the  proper  length  of  vF  is  easily  computed ;  and  similarly, 
using  T'  and  the  triangle  mvGr,  we  calculate  mCr.  Joining  v  and 
m,  the  required  right  line  vm  is  determined. 

107.  Three-point  Construction.  Equilibrium  Polygon  for  Non- 
vertical  Forces.  Preliminary  Step. — In  the  construction  and  proof 
of  §  378a,  M.  of  E.,  it  is  supposed  at  the  outset  that  the  action- 
line  of  the  resultant  (/?,)  of  all  the  forces  acting  between  points 
A  and  p  has  been  found ;  similarly,  that  of  the  resultant  (/2a)  of 
the  forces  acting  between  p  and  B  ;  and  that  of  the  resultant  (R) 
of  the  two  partial  resultants. 

It  is  here  intended  to  give  the  detail  of  finding  these  three 
lines  and  to  make  clearer  the  scope  and  intent  of  the  problem. 
Fig.  128.  Let  A^p^  and  B  be  the  three  points  through  which 
the  equilibrium  polygon  is  to  pass,  and  1,  2,  etc.  (to  6  inclusive) 
(on  left  of  figure),  the  given  forces  in  magnitude  and  position  ; 
1,  2,  and  3  acting  between  A  and  JP;  and  4,  5,  and  6,  between  p 
and  B.  Lay  off  the  "  load-line"  8TU,  on  some  convenient  scale  of 
force,  and  select  any  pole  as  0'".  Draw  the  "  rays,"  0'"8^  etc. ; 
and  also  lines  parallel  to  them,  in  proper  order,  beginning  at  a 
convenient  point  A"  (not  ^4,  necessarily),  so  as  to  form  an  equilib- 
rium polygon,  A" B" ,  as  shown.  The  first  segment  is  parallel 
to  0"'S ;  the  last,  to  0'"  U ;  and  the  segment  connecting  forces 
3  and  4  (in  this  particular  instance),  parallel  to  O'"T.  The  inter- 
section of  the  first  and  last  segments  gives  Mfl ',  a  point  in  the 


118 


NOTES   AND   EXAMPLES   IN   MECHANICS. 


action-line  of  R,  the  resultant  of  all  the  given  forces,  1  .  .  6 ;  and 
similarly,  the  other  intersections  N"  and  0"  are  points  in  the 
action-lines  of  Rl  and  R^,  respectively.  Right  lines  through 
these  three  points,  parallel  respectively  to  SU,  ST,  and  TU, 

M' 

Preliminary  Step;  Equil., 
Polygon  thro'  Three  Points; 
Forces  not  Vertical. 


FIG.  128. 

should  meet  in  a  common  point  €' ',  and  are  the  respective  action- 
lines  desired. 

The  further  steps  are  those  given  on  pp.  460  and  461,  M.  of 
E.  Our  present  equilibrium  polygon,  A"  B" ,  is  of  no  further 
use  ;  but  the  load-line  STU  will  still  serve,  after  the  lines  Mf A 
and  M ' B  have  been  located  according  to  §  378a.  We  can  then 
draw  through  S  and  U  parallels  to  Mf A  and  MB,  respectively, 
and  by  the  intersection  of  these  parallels  with  each  other  deter- 
mine the  pole  corresponding  to  the  final  equilibrium  polygon 
which  is  to  pass  through  A9  p,  and  B. 

It  is  to  be  noted  that  the  problem  of  the  "  Shear-legs"  of  §  59 
of  these  Notes,  and  also  Problem  2  (of  the  two  links),  p.  35,  M. 
of  E.,  are  cases  of  a  three-hinged  arch-rib,  and  can  be  treated 
graphically  in  the  same  manner  ;  and  thus  the  "  special"  equilib- 
rium polygon  and  its  corresponding  pole  and  rays  (i.e.,  force- 
diagram)  determinedo 

The  ray  parallel  to  the  segment  passing  through  the  inter- 
mediate joint  (p)  gives  the  amount  and  direction  of  the  pressure 
on  the  hinge  of  that  joint ;  and  corresponding  statements  may  be 
made  for  the  two  extreme  joints. 


CHAPTER  VIII. 

MISCELLANEOUS  NOTES. 

108.  Co-ordinates  of  Centre  of  Gravity  —  Fuller  Explanation.  — 
Assume  the  various  small  particles  of  a  rigid  body  to  be  num- 
bered 1,  2,  3,  etc.,  and  call  their  respective  volumes  d  Vl  ,  d  F,  , 
d  F,  ,  etc.  (cubic  feet),  and  their  #-co-ordinates  xl  ,  a?2  ,  xz  ,  etc. 
(feet).  If  the  body  is  heterogeneous,  the  particles  may  be  of  dif- 
ferent densities;  for  example,  particle  1  may  be  of  such  a  density 
that  a  cubic  foot  of  material  of  that  density  would  weigh  100  Ibs., 
while  a  cubic  foot  of  the  material  of  which  particle  2  is  composed 
would  weigh  110  Ibs.  ;  and  so  on.  Or,  in  symbols,  the  "  heavi- 
ness," or  rate  of  weight,  of  particle  1  is  yl  =  100  Ibs.  per  cubic 
foot,  while  that  of  particle  2  is  y9  =  110  Ibs.  per  cubic  foot.  A 
similar  notation  would  apply  to  all  the  other  particles. 

The  respective  weights,  then,  of  the  particles  (or  force  of  the 
earth's   attraction    on   them)    are   y^  F,,    7^F2,    ysdV3,   etc. 
(pounds),  and  if  we  substitute  these  for  the  forces  Pt  ,  Pa  ,  P3 
etc.,  in  the  expression  for  the  x  of  the  centre  of  parallel  forces 
(foot  of  p.  16,  M.  of  E.),  we  obtain 


_  w&  V*  +  wA  V       **     .     .  .  . 
r,dV,  +  r,d^  +  r3dVa+... 

which  in  the  compact  notation  of  calculus,  the  particles  being 
taken  "  infinitely  small  "  and,  therefore,  "  infinite  "  in  number, 
can  be  written 

fxydV 

or    m^-mfdV   .    .  (b) 


JydV 

where  C  is  the  total  weight  of  the  body,  =    /  yd  V. 

If  the  body  is  homogeneous,  all  the  particles  have  a  common 
"  heaviness,"  which  we  may  call.y»  and  factor  out,  thus  obtaining 

119 


120  NOTES   AND   EXAMPLES   IN   MECHANICS. 

y '  m 


°r     "- 


Ym 

from  which  the  ym  can  be  cancelled,  leaving 

_       1    /* 

x=— rl  xdV,       .     .    .    «    .    '.'•  *      (c) 

T/t/  ^ 

where  V  denotes  the  total  volume  of  the  body.  (Note  that  the 
factoring  out  of  a  common  multiplier  from  a  parenthesis  corre- 
sponds to  taking  a  constant  outside  of  the  integral  sign.) 

109.  The  Time-velocity  Curve  and  its  Use.— From  eq.  (I.),  p.  50, 
M.  of  E.,  we  have  ds  -r-  dt  =  v,  the  velocity  of  a  moving  point 
at  any  instant.  Hence,  also,  ds  =  vdt,  or  the  element  of  dis- 
tance equals  the  product  of  the  velocity  at  that  instant  by  the 
element  of  time.  If  now,  whatever  the  character  of  the  recti- 
linear motion,  we  conceive  a  curve  to  be  plotted,  in  which  the 
time  (from  some  initial  instant)  is  laid  off  as  an  abscissa,  and 
the  velocity  of  the  moving  point  as  an  ordinate,  this  curve 
may  be  called  a  u  time-velocity"  curve  for  the  particular  kind  of 
motion,  being  different  for  different  kinds  of  motion.  (Of 
course,  proper  scales  must  be  selected  in  laying  off  distances  on 
the  paper  to  represent  the  quantities  time  and  velocity.) 

The  equation  expressing  the  relation  between  the  two  vari- 
ables, time  and  velocity,  may  be  regarded  as  the  equation  to  the 
curve.  Thus,  in  uniform  motion  we  have  the  velocity  v  con- 
stant, so  that  the  curve  is  a  right  line  parallel  to  the  horizontal 
axis,  or  axis  of  time,  as  AL  in  Fig.  130,  where  OA  represents  the 
constant  velocity.  If  the  motion  is  uniformly  accelerated,  that 
is,  has  a  constant  acceleration,  we  have  v  =  v0  -\-pt  [where  p  is 
the  acceleration,  or  rate  of  change  of  the  velocity,  and  v0  is  the 
initial  velocity  (for  t  =  0)],  and  note  that  the  quantity  pt,  or 
total  gain  in  velocity  over  the  initial  velocity  v0,  is  directly  pro- 
portional to  the  time,  so  that  the  curve  obtained  is  a  straight  line 
inclined  to  the  axis  of  abscissas,  as  BH  in  Fig.  130,  BO  repre- 
senting the  initial  velocity  VQ  . 

In  general,  let  KVM,  Fig.  129,  be  the  time- velocity  curve 


MISCELLANEOUS   NOTES. 


121 


M/ 


l  = 


for  any  rectilinear  motion.     Here  we  note  that  the  product  v  .  dt 

at  any  instant  during  the  motion  is  represented  by  the  area  of  the 

vertical  strip  "FT?,  whose  width  is  dt  and  length  v  (by  scale). 

Hence  the  element  of  distance  ds  (feet)  is  proportional  to  that 

area,  and  the  whole  distance  (s)  described  from  the  beginning  is 

represented  by  the  sum  of  the  areas  of  all  such  strips  from  OK 

to  FT?,  i.e.,  by  the  area  OKVR\   while  the  distance  (sn)  de- 

scribed  between  t  =  0   and  t  =  tn  will  be  represented  by  the 

complete  area  OKVMN. 

If  now  ON  be  regarded  as  the  base  of  the  figure  OKMN,  we 

may  conceive  of  a  rectangular  figure  OALN  having  the  same 

area  and  same  base  as  OKVMN,  and  having 

some  altitude  A  0  which  can  be  looked  upon 

as  representing  the  "average  velocity  "of  the 

motion  between  t  —  0  and  t  =  tn.   By  "  aver- 

age velocity  "  would  be  meant  that  constant 

velocity  necessary  in  a  uniform,  motion  to 

enable  a  moving  point  to  describe  the  dis- 

tance sn  in  the  same  time  tn  as  in  the  ac- 

tual motion.     In  other  words,  the  "  aver- 

age velocity  "  is  the  result  obtained  by  di- 

viding  the  whole  distance  by  the  whole  time,  and  is  represented 

by  the  altitude  AO,  since  the  area  of  a  rectangle  is  equal  to  the 

product  of  its  base,  ON(=  tn\  by  its  altitude,  AO. 

As   a   useful   instance   consider  again  a   uniformly   acceler- 

ated motion.  Its  time-  velocity  curve  is  a  straight  line,  BH, 
Fig.  130,  the  initial  velocity  v9  being  rep- 
resented by  QB,  and  the  final,  vn  (for 
t  =  in),  by  HN.  The  distance  described  is 
represented  by  the  area  of  the  trapezoid 
OBHN.  This  area  is  equal  to  that  of  a 
rectangle  of  the  same  base  ON  (i.e.,  tn) 
and  of  an  altitude  OA  =  half  the  sum  of 
OB  and  NH\  i.e.,  sn  =  %(v0  +  vn)tn.  In 

other  words,  the  average  velocity  between  t  =  0  and  t  =  tn  for 

the  uniformly  accelerated  motion  is  J(-yQ  +  vn). 


time 


FlG- 


vn 


FIG.  130. 


122 


NOTES   AND    EXAMPLES   IN    MECHANICS. 


FIG.  131. 


If  the  initial  velocity  of  the  uniformly  accelerated  motion  is 
zero,  the  time-velocity  curve  becomes  a  straight  line  passing 
/  through  the  origin,  0,  viz.,  OH,  Fig.  131  ; 
and  in  that  case  the  distance  sn  is  repre- 
sented by  the  area  of  the  triangle  OHN, 
and  the  average  velocity  OA  is  one  half  the 
final,  or  the  final  is  double  the  average. 
Therefore  to  obtain  the  whole  distance  de- 
scribed we  must  multiply  the  whole  time  by 
one  half  the  final  velocity  ;  i.e.,  sn  =  %vntn. 
If,  then,  in  this  case  of  uniformly  ac- 
celerated motion  with  initial  velocity  =  zero,  we  divide  the  whole 
distance  by  the  whole  time,  we  do  not  obtain  the  final  velocity  (a 
common  error  with  students),  but  only  the  "  average  velocity," 
in  the  sense  defined  above.  This  result  must  then  be  doubled 
to  obtain  the  final  velocity. 

As  an  instance  where  the  average  velocity  is  one  third  of  the 
final  (the  initial  velocity  being  zero),  consider  the  case  of  variably 
accelerated  motion  represented  by  the  relation  v  =  qf,  where 
q  is  a  constant.  The  time-velocity  curve  will  have  the  form 
OPH,  Fig.  131,  a  parabola  with  vertex  at  0.  The  area  OPEN 
will  be  one  third  of  that  of  the  circumscribing  rectangle,  and 


hence  is  equal  to  O      X  one  third  of  NH,  i.e.,  to  ON  X 
Hence  OB,  or  one  third   of  the  final  velocity,  is  the   average 
velocity. 

Or,  mathematically,  in  detail,  ds,  =  vdt,  =  qfdt  ; 
therefore 


sn  = 


But  for  t  —  tn  we  have  v,  =  vn,  =  eft*,  and  hence 


(1) 


That  is,  the  average  velocity  is  equal  to  one  third  of  the  final. 

110.  Reduction- Formulae  for  Moment  of  Inertia  of  a  Plane 
Figure. — (To  replace  §  88,  M.  of  E.,  as  regards  the  Moment  of 
Inertia  of  a  plane  figure.) 


MISCELLANEOUS   NOTES. 


123 


Definition.  —  Any  right  line  containing  the  centre  of  gravity 
ot  a  plane  figure  is  called  a  "  gravity  axis  "  of  that  figure. 

Theorem.  —  The  moment  of  inertia  of  a  plane  figure  about 
a  given  axis  in  its  own  plane  is  equal  to  its  moment  of  inertia 
about  a  gravity  axis  parallel  to  the  given 
axis,  augmented  by  the  product  of  the  area 
of  the  figure  by  the  square  of  the  distance 
between  the  two  axes. 

Proof.—  Fig.  132.     Let  dF  be  the  area 
of  any  element  of  the  plane  figure,  and 
the  distance  of  that  element  from  any  axis 
X  in  the  plane  of  the  figure  ;  while  z  is  its 
distance  from  a  "  gravity  axis,"  g,  parallel  ~ 
to  axis  X.      By  definition  we  have  Ix  = 


Fia 


axes. 


\  but  z'  =  z  + 
Hence 


being  the  distance  between  the  two 


dF. 


Now,  from  the  theory  of  the  centre  of  gravity,  we  have  (see 
eq.  (4),  p.  19,  M.  of  E.)fsdF=  F~z,  where  "z  is  the  distance  of 
the  centre  of  gravity  C  from  the  axis  g.  But  g  is  a  gravity  axis, 
so  that  -5  —  0  ;  and  hence  fadF  =  0.  Also  fdF  =  F,  the  whole 
area;  %ud.fzdF=Ig.  Whence,  finally, 

Ix  =  Ig  +  Fd\     ...    (4)    Q.  E.  D. 
It  also  follows,  by  transposition,  that 

Ig  =  l.-F#\     .    ...    .    .    (4a) 

which  shows  that  the  moment  of  inertia  of  a  plane  figure  about 
a  gravity  axis  is  smaller  than  that  about  any  other  axis  parallel 
to  that  gravity  axis. 

The  moment  of  inertia  of  a  plane  figure  plays  an  important 
part  in  the  theory  of  beams  subjected  to  bending  action,  the 
transverse  section  of  the  beam  forming  the  plane  figure  in  ques- 
tion ;  somewhat  as  the  mere  area  of  the  section  does  when  the 
beam  or  rod  is  subjected  to  a  straight  pull. 


124  NOTES   AND   EXAMPLES   IN   MECHANICS. 

111.  Miscellaneous  Examples.     (See  opposite  pages  for  figures.) 

1.  Fig.  A.  Given  the  data  of  the  figure,  find  the  stress  in  every  two-force 
piece,  and  the  three  pressures  exerted  on  the  pin  at  J5. 

2.  Wind  from  the  S.W.,  30  miles  per  hour.     Ship  going  toward  the  N.W. 
at  10  miles  per  hour.     At  what  angle  with  ship's  course  should  a  vertical  sail 
be  placed  that  the  air- par  tides  may  strike  it  at  an  angle  of  30°  on  the  hinder 
side  ? 

3.  Locate,  by  calculus,  the  centre  of  gravit}'  of  the  plane  figure  in  Fig.  B, 
the  equation  to  the  upper  bounding  curve  being  xy  =  20  sq.  ft. 

4.  In  the  rectilinear  motion  of  a  material  point  weighing  12  Ibs.,  and  mov- 
ing horizontally  on  a  rough  surface,  friction  from  which  is  the  only  horizon- 
tal force,  we  note  that  positions  A,  B,  C,  and  D  are  passed  at  the  following 
times  (by  the  clock),  respectively  : 

3 h.  4  m.  8.1239  sec.;  3  h.  4  m.  8.2350  sec.;  3  h.  4  m.  8.3490  sec.; 
3  h.  4  m.  8.4658  sec. 

The  distance  from  A  to  B  is  10.00  ft. ;  from  B  to  G,  10.02  ft. ;  and  C  to  D, 
10.05  ft. 

Find  (approx'ly)  the  acceleration  of  the  motion  as  the  material  point 
passes  the  position  B ;  also  for  C;  and  what  must  be  the  value,  in  pounds,  of 
the  friction  at  B  ? 

5.  Fig.  C.  Find  the  stress  in  each  two-force  piece,  and  pressure  on  pin  at 
A  ;  also  pressure  of  pin  B  on  bar  BD. 

6.  Of  the  solid  right  cylinder  in  Fig.  D,  the  internal  conical  portion  is  of 
lead,  whose  spec.  grav.  is  11.3,  while  the  remainder  is  of  cast  iron.     Find  the 
centre  of  gravity  of  the  whole  solid. 

7.  In  Fig.  80  on  p.  67,  Notes,  compute  the  pressure  between  the  block  and 
the  guide  when  the  former,  having  started  from  rest  at  A,  is  passing  position 
B,  45°  from  A. 

8.  Fig.  E.  The  block  weighs  40  Ibs.     The  cord  d  .  .  .  e  is  attached  to  it 
at  angle  15°  with  the  plane,  which  is  smooth.     If  the  block  is  to  be  permitted 
to  slide  from  rest  down  the  plane,  under  action  of  gravity,  the  cord,   and 
the  plane,  what  constant  tension  (Ibs.)  must  be  maintained  in  the  cord  that  a 
velocity  of  12  ft.  per  sec.  may  be  generated  in  2  seconds  ?    Afterwards,  what 
new  value  must  be  given  to  this  tension  if  the  velocity  is  to  continue  at  that 
figure  (12  ft.  per  sec.)  ? 

9.  In  the  vertical  fall  of  a  material  point,  certain  consecutive  small  space- 
intervals  are  given,  =  a,  b,  c,  and  d ;  and  the  corresponding  time-intervals, 
/,  t',  t",  t"r.     Derive  approximate  formulae  for  the  velocities  at  mid-points  of 
these  time-intervals  ;  the  accelerations  near  the  end  of  some  of  these  intervals  ; 
and  the  corresponding  resultant  force  that  must  be  acting,  the  weight  of  the 
body  being  O. 

10.  Fig.  F.  Find  the  amount  and  position  of  the  pressure  between  the  bar 
AB  and  each  of  the  four  pins  passing  through  it ;  also  the  stress  in  DB. 


t 


^Xl<  I 

VH C-< J 

I 


400  Z&s. 


B 


1200  Z&s. 


Pi  ~ 

Ui--— - 
J_ 


12" 


20  Ibs.         10  »*• 


R 


t 
¥ 

L. 


\12lbs. 


[A  is  inside 


30' 


N.B.  The  small  pulley  is  free 
on  the  axle  of  wheel  B . 


chain 


2400  Ibs. 


20' 


JV 


'-—friction  * 
12000  Ibs. 


126  NOTES   AND    EXAMPLES   IN   MECHANICS. 

11.  Fig.  G.  The  two  pulleys  run  on  fixed  bearings,  without  friction.     If 
the  given  three  weights  be  allowed  to  come  to  a  position  of  equilibrium,  the 
knot  A  being  a  fixed  knot,  find  the  angles  the  two  oblique  cords  make  with  the 
vertical,  respectively. 

12.  Find  the  moment  of  inertia  about  the  axis  X  of  the  symmetrical 
plane  figure  shown  in  Fig.  H.     The  equation  to  the  bounding  curve  AB  is 
(x  —  a'}  -r-  (a  —  a')  =  z* -+-  h*. 

13.  Fig.  J.  The  block  weighs  12  Ibs.,  and  in  sliding  down  the  rough  in- 
clined plane  encounters  a  variable  friction,  which  in  Ibs.  —  3  times  the  dis- 
tance, s,  from  the  starting  point,  in  feet.     It  starts  from  rest  at  A.     Find  the 
velocity  acquired  on  its  reaching  a  position  2  ft.  vertically  below  A. 

14.  Compute  the  moment  of  inertia  of  the  plane  figure  in  Fig.  K  about  the 
axis  X ;  also  the  corresponding  radius  of  gyration. 

15.  Compute  the  y  co-ord.  of  the  centre  of  gravity  of  the  plane  figure 
shown  in  Fig.  L.     The  equation  to  bounding  curve  is  a;3  =  4y,  with  the  foot 
as  linear  unit. 

16.  A  block  of  50  Ibs.  weight  is  started  along  a  rough  horizontal  table  with 
an  (initial)  velocity  of  40  ft.  per  sec.     If  the  friction  met  with  is  variable,  and, 
in  Ibs.,  equal  to  700  times  the  body's  weight  -s-  (1200  +  the  square  of  the 
veloc.  in  ft.  per  sec.),  find  the  time  and  distance  in  which  the  body  comes  to 
rest. 

17.  Fig.  M.  Find  the  tensions  in  all  chains  and  the  pressure  on  pin  A  and 
under  wheel  B. 

18.  Fig.  N.  Find  the  moment  of  inertia  of  the  plane  figure  about  the  grav- 
ity-axis X  which  is  perpendicular  to  the  axis  of  symmetry. 

19.  Fig.  P.  The  ram  A,  of  1200  Ibs.  weight,  falls  from  rest  through  20 
ft.,  and  has  then  an  inelastic  impact  with  the  pile  B,  of  weight  400  Ibs.     Com- 
pute their  common  velocity  after  the  impact,  the  Kinetic  Energy  lost  in  the 
impact,  and  the  distance  the  two  bodies  will  sink  after  the  impact,  overcom- 
ing the  constant  frictional  resistance  of  12000  Ibs.  on  side  of  hole. 

20.  Fig.  Q.  Block  of  10  Ibs.  weight  on  inclined  plane.     It  starts  from  rest 
at  A.     If  friction  on  AB  is  2  Ibs.,  while  on  EG  it  is  3  Ibs.,  compute  the  time 
of  reaching  position  G,  and  its  velocity  at  that  instant. 

21.  Fig.  R.  A  block  of  12  Ibs.  weight  falls  from  rest,  freely  through  the 
first  7  ft.,  but  then  strikes  the  head  of  spring  A,  which  opposes  a  resisting 
force  at  rate  of  100  Ibs.  per  inch  of  shortening  ;  and  3  inches  further  down 
Strikes  spring  B,  offering  160  Ibs.  of  resistance  per  inch  of  shortening.     Where 
is  the  block  when  (momentarily)  brought  to  rest  (supposing  the  elastic  limit  of 
the  springs  not  passed)  ? 

22.  A  bod}r  weighing  12  Ibs.  is  given  an  initial  upward  vertical  velocity  of 
10  ft.  per  sec.,  being  thereafter  acted  on  only  by  gravity  and  a  variable  hori- 
zontal force  =  [TV  of  time  in  sec.]  Ibs.     Find  the  equation  to  its  path. 


B 


f« 5^ 


1200  Ibs. 


10-- — 


U- 


'   — ~"--^. 


22  Ibs. 


R 


m  12  Ibs. 


U>    ' 

b 


The  small  pulley  is  free 
on  the  axle  of  wheel  B. 


2400  tes. 


20' 


'//     12000  Ibs. 


128  NOTES    AND   EXAMPLES    IN   MECHANICS. 

112.  Answers  to  Preceding  Problems.—  1.  339.1  Ibs.  and  240 
Ibs.  ;  on  pin  A,  240  Ibs.  vertically,  288.3  Ibs.  at  33°  42'  with  hor- 
izontal, and  466.2  Ibs.  at  30°  56'  with  vertical. 

2.  41°  34'  ;  or  sail  pointing  3°  26'  K  of  West. 

3.  so  =  3.28  ft.  ;  y  =  3.28  ft.  ;  area  of  figure  =  18.31  sq.  ft. 

4.  In  passing  B,  ace.  =  18.72  ft.  per  sec.  per  sec.  ;  C,  16.05; 
friction  as  the  material  point  passes  B,  6.98  Ibs. 

5.  154  Ibs.  compression,  and  1200  Ibs.  tension  ;  at  A,  pressure 
=  738  Ibs.  at  25°  with  horiz.  ;  pressure  of  pin  B  on  bar  BD  = 
1132  Ibs.  at  36°  45'  with  vertical. 

6.  x  =  5.39  inches  from  left-hand  base. 

7.  Pressure  is  42.44  Ibs. 

8.  First  value  of  tension,  18.92  Ibs.  ;  second,  26.63  Ibs. 

9.  Near   the   end   of   the   first   interval    the   acceleration  = 


—  at'}     ,  ,  ,     .  ,  .   ,        ,   2(X  —  W 

that  near  end  of  second  interval, 


+      . 

10.  Pressures  are  1600,  45T?   2263,  1143  Ibs.,  respectively. 
There  is  no  stress  in  piece  DB. 

11.  On  the  left,  87°  43'  ;  on  the  right,  65°  IT. 

12.  The  moment  of  inertia,  about  X,  is  -f  [aA3  -f-  \a'h  ']. 

13.  The  velocity  acquired  is  7.13  ft.  per  second. 

14.  Ix  =  330.66  biquad.  inches;  rad.  gyr.  =  3.02  inches. 

15.  The  area  is  80  sq.  ft.  and  y  =  15.61  ft. 

16.  Time  —  3.075  sec.,  to  come  to  rest.     Distance  —70.9  ft. 

17.  Tension  in  upper  chain,  5838  Ibs.  ;  in  the  lower  oblique 
chain,  2770  Ibs.     Pressure  under  wheel  B,  1385  Ibs.     Pressure 
on  pin  at  A  is  7503  Ibs.,  and  is  directed  toward  the  right  at  an 
angle  of  6°  37'  above  the  horizontal. 

18.  Distance  of   centre  of  gravity  from  the  horizontal  line 
drawn  through  the  upper  corners  is  2.20  inches.     Moment  of 
inertia  about  the  gravity  axis  X  is  8.12  biquad.  inches. 

19.  26.91  ft.  per  sec.  ;  6000  ft.-lbs.  lost;  1.73  ft.  sinking  after 
impact. 

20.  Time,  A  to  (7,  =  2.06  sec.;  velocity  at  #=18.05  ft.  per  sec. 

21.  The  block  is  7  ft.  4.297  inches  below  its  starting-point, 
i.e.,  has  shortened  spring  A  an  amount  of  4.297  inches. 


PKOBLEMS    AND    EXAMPLES. 


129 


(< F — H 


FIG.  150. 


113.  Examples. — 1.   Fig.  149.   The  hollow  shaft  A  is  to  be 

twice  as  strong  torsionally  (i.  e. ,  as  to  tor-  ^v ^ 

sional  moment)  as  the  solid  shaft  B ;  while 
the  material  of  A  is  only  half  as  strong  as 
that  of  B.  Given  the  lengths  Z  and  Z", 

O  i  "•  (,  ~  ~~-n 

the  radius  r"  of  j5,  and  the  outer  radius,  FIG.  149. 

r,  of  ^1,  determine  a  proper  value  for  r',  the  inner  radius  of  A, 

for  above  conditions. 

2.  Fig.    150.   Homogeneous    prismatic   beam;     rectangular 
section ;   width  =  h  and  height  =  h ;   placed  with  h  horizontal, 

and  loaded  uniformly  over  its  whole 
length  at  rate  of  w  Ibs.  per  running 
inch.  Given  the  whole  length,  Z, 
and  position  of  support  A  (at  left 
end),  where  (i.e.,  distance  a  =  ?) 
shall  we  place  support  B  that  the 
moment  of  stress-couple  in  section  over  B  shall  equal  (without 
regard  to  sign)  the  greatest  moment  of  stress- couple  occurring 
between  A  and  the  point  of  inflection,  F,  of  the  elastic  curve  ? 

Also,  after  a  is  found,  find  the  maximum  shear,      (b  and  h 
are  given,  aiid'also  w\  elastic  limit  supposed  not  passed.) 

3.  Fig.  151.   By  the  principles  of  the  graphical  statics  of 
mechanism,  having  the  resisting   force    Q    given    in 
amount  and  position,  given  also  all  friction  angles  con- 
cerned, and  considering  all  kinds  of  frictional  action, 

find  the  value  of  P  for  forward  motion ;  also  for  back- 
ward    motion ;     the 
efficiency    of     the 
mechanism. 

4.  Fig.  152.   The 
load,   of    weight    P 
=  60  Ibs.,  is  gradu- 
ally applied  at  lower 

end  of  compound  vertical  round  wire  (of  wrought  iron)  whose 
upper  end  is  fixed.  Neglecting  the  weight  of  the  wire,  compute 
the  total  elongation  of  the  wire  and  the  work  done  on  the  wire 


130  NOTES   AND   EXAMPLES  IN  MECHANICS. 

during  the  gradual  stretching.     (Lengths  are  100'  and  120'; 
diameters  J  in.  and  T^-  in.,  respectively.) 

5.  Fig.  153.  The  horizontal  prismatic  beam  is  of  rectangular 
section ;  width  J,  horizontal.  Its  height  h  is  to  be  three  tim-es 
its  width.  Beam  of  timber.  The  length  and  character  and 
amount  of  loading  are  given  in  the  figure,  the  weight  of  the 
beam  itself  being  neglected. 

VSSS»»S%£%»S;;S/S/A  There  being  three  (local)  maxi- 
AIHP           HHc 

fxx^xjxxxxxxxx     ~.  mum  moments  (of  stress-couple), 
"*       m"~n  viz-> at  ^' B* and  ^ locate  sec~ 
^ ku2*oii«  ^on    -B  by   determining    distance 
i 


FIG.  153.  u#";    compute   the   moments   at 

A^  B,  and  (7;  and  then  determine  the  minimum  safe  dimen- 
sions to  be  given  to  the  section  of  the  beam.  (The  load  of  five 
tons  is  uniformly  distributed  along  the  ten  feet.) 

6.  Fig.  154.   The  short  vertical  cylindrical  body   abed  is 
fixed  at  the  upper  end  while  sustaining  at  its  lower 
end  a  weight  of  4000  Ibs.,  whose  line   of   action 
prolonged    upward    passes    through   the    extreme 
edge,  &,  of  the  section  db. 

The  horizontal  section  of  the  body  is  a  circle 
of  radius  =  0.  5  inch.  It  is  required  to  compute 
the  stress  per  square  inch  at  point  a  of  the  section 
db,  and  also  that  at  point  b. 

Also,  what  would  these  stresses  be  if  the  line 
of  action  of  the  4000  Ibs.  load  passed  through  the 
centre  of  the  section  ab  ? 

114.  Answers  to  Problems  in  §  113.    (For  6,  see  below.*) 


1.  rf  =     r4  -  4rr"\ 

2.  a  =  (1  —  4/i)Z,  =  0.293?.     The  maximum  shear  is  just 
on  the  left  of  support  B  and  is  Jm  —  (  V2  —  V)wl  =  0.414i0Z. 

3.  (Solved  graphically.) 

4.  Total    elongation  =  1.21    inches.       Work    done  =  36.3 
in.  -Ibs. 

5.  Distance  a  =  4.4  ft.     Moments  are  1.00,  3.84,  and  4.00 

*  Answer  to  Ex.  6  :  —  At  a,  12.7  tons  per  sq.  in.  tension  ;  at  &,  7.63  tons  per 
sq.  in.  compression. 


THE    "  IMAGINARY   SYSTEM"    IN   ROTARY   MOTION.     131 

ft. -tons,  respectively.  Taking  Br  =  1000  Ibs.  per  sq0  in.,  we 
have  h  =  12,  and  b  =  4,  in. 

115.  The  "Imaginary  System." — In  conceiving  of  the  imagi- 
nary equivalent  system  in  §  108,  M.  of  E.,  applied  to  the 
material  pointo  supposed  destitute  of  mutual  action,  and  not 
exposed  to  gravitation,  we  employ  the  simplest  system  of  forces 
that  is  capable,  by  the  Mechanics  of  a  Material  Point,  of  pro- 
ducing the  motion  which  the  particles  actually  have.  If  now  the 
mutual  actions,  coherence,  etc.,  were  suddenly  re- established, 
there  would  evidently  be  no  change  in  the  motion  of  the  assem- 
blage of  particles ;  that  is,  in  what  is  now  a  rigid  body  again ; 
hence  the  imaginary  system  is  equivalent  to  the  actual  system. 

In  applying  this  logic  to  the  motion  of  translation  of  a  rigid 
body  (see  §  109  and  Fig.  122,  M.  of  E.)  we  reason  as  follows: 

If  the  particles  or  elementary  masses  did  not  cohere  together, 
being  altogether  without  mutual  action  and  not  subjected  to  grav- 
itation, their  actual  rectilinear  motion  in  parallel  lines,  each  hav- 
ing at  a  given  instant  the  same  velocity  and  also  the  same  accel- 
eration, p,  as  any  other,  could  be  maintained  only  by  the  appli- 
cation, to  each  particle,  of  a  force  having  a  value  ==  its  mass  X  p, 
directed  in  the  line  of  motion.  In  this  way  system  (II.)  is  con- 
ceived to  be  formed  and  is  evidently  composed  of  parallel  forces  all 
pointing  one  way,  whose  resultant  must  be  equal  to  their  sum,  viz. 

/  dM  X  p>    But  since  at  this  instant  p  is  common  to  the  motion 

of  all  the  particles,  this  sum  can  be  written  p  I  dM,  =  the  whole 

mass  M  X  p- 

If  now  the  mutual  coherence  of  contiguous  particles  were  sud- 
denly to  be  restored,  system  (II.)  still  acting,  the  motion  of  the 
assemblage  of  particles  would  not  be  affected  (precisely  as  the  fall- 
ing motion  in  vacua  of  two  wooden  blocks  in  contact  is  just  the 
same  whether  they  are  glued  together  or  not)  and  consequently  we 
argue  that  the  imaginary  system  (II.),  is  the  equivalent  of  what- 
ever system  of  forces  the  body  is  actually  subjected  to,  viz.  sys- 
tem (I.),  (in  which  the  body's  own  weight  belongs)  producing  the 
actual  motion. 


132  NOTES   AND  EXAMPLES  IN   MECHANICS. 

Since  the  resultant  of  system  (II.)  is  a  single  force,  = 
parallel  to  the  direction  of  the  acceleration,  it  follows  that  the 
resultant  of  the  actual  system  is  the  same. 

116.  Angular  Quantities— Rotary  Motion  about  a  Fixed  Axis. 
— In  Fig.  155  let  the  body  MLN  rotate  about  the  fixed  axis  C 
(perpendicular  to  paper),  the  initial  posi- 
tion of  the  arm  CL  having  been  CY. 
Now  the  angular  motion  of  the  whole 
rigid  body  is  the  same  as  that  of  the  arm 
CL.  If  in  a  small  time- interval,  dt,  the 
arm  passes  from  position  CA  to  position 
CB,  and  thus  describes  the  small  angle 
ACB,  whose  value  in  ?r-measure  is  da 
radians,  the  angular  velocity  at  about  the  mid-point  of  angle 
A  CB  is  GO  =  da  -r-  dt.  In  the  next  and  equal  time-interval  a 
slightly  different  angle,  da'  radians,  is  described ;  and  if  in  the 
figure  we  lay  off  angle  B  CE  equal  to  A  CB,  the  angle  ECB,  or 
difference  between  da  and  da  may  be  called  d*a.  And  so  on, 
for  any  number  of  consecutive  (dor)'s,  described  in  equal  times, 
each  =  dt.  If  the  motion  is  uniform,  all  the  (dcfy's  are  equal ; 
that  is,  the  angular  velocity  is  constant,  each  d*a  being  =  zero. 
If  all  the  (d*ay&  are  equal,  the  motion  is  uniformly  accel- 
erated, the  angular  acceleration,  0,  being  thus  determined :  The 
gain,  or  change,  of  angular  velocity  occurring  between  the  mid- 
point of  ACB  and  that  of  BCD  is  -j-  -  -^,  or  GO'  —  oo\  and 

if  this  be  divided  by  the  time,  dt,  occupied  in  acquiring  the 
gain,  we  have  for  the  rate  of  change  of  angular  velocity,  that 

GO1  —    GO  d(&> 

is,  for  the  angular  acceleration,  the  value  6  -    —^- —     -  ^-. 
Another   form   is   this:    if   the   gain   of   angular  velocity, 

— -^,  be  divided  by  dt,  we  have 

dt       dt 

da'  -da 


If  the  successive  (^V)'s  are  unequal  (the  successive  (da)^ 


ANGULAR   QUANTITIES.       ROTAEY    MOTION.  133 

being  described  in  equal  time-intervals,  it  must  be  remembered), 
the  motion  is  some  kind  of  variably  accelerated  angular  motion ; 
e.g.,  in  a  harmonic  (oscillatory)  rotary  motion  the  angular 
acceleration  at  any  instant  is  directly  proportional  to  the  angular 
space  between  the  ' '  arm ' '  or  reference  line  of  the  body  and  the 
middle  of  its  oscillation,  and  is  of  contrary  sign;  i.e.,  6=  —Aa, 
where  A  is  a  constant. 

Example  1.  At  a  certain  part  of  its  revolution  a  fly-wheel 
is  found  to  describe  just  1°  in  0.01  second.  Here  we  have 
da=  0.01745  radian,  and  dividing  this  by  the  dt,=  0.01  sec., 
we  obtain  GO  =  1.745  radians  per  second  as  the  angular  velocity 
of  the  wheel  at  this  part  of  its  progress,  as  nearly  as  the  data 
permit.  (Strictly,  this  value  of  GJ  is  only  the  average  value  of 
the  angular  velocity  for  this  small  but  finite  portion  of  the 
motion.  The  data  are  insufficient  to  determine  whether  the 
velocity  is  variable  or  constant.) 

Example  2.  The  same  wheel,  besides  describing  1°  in  0.010 
sec.,  is  found  to  describe  1°  2'  in  the  next  0.01  second.  Com- 
pute the  angular  acceleration,  as  near  as  may  be  from  these  data, 
for  this  part  of  the  motion.  As  before,  da  =  0.01745  radian, 
and  we  now  have  the  additional  fact  that  da'  =  0.01803  radian, 
each  of  the  time-intervals  being  dt  =  0.010  sec.  If  we  substi- 
tute directly  in  eq.  (VII),  there  results 

0.000581  radian 

6  =  -  — =  5. 81  rad.  per  sec.  per  sec. 

(0.01  sec.)' 

That  is  to  say,  at  this  part  of  the  wheel's  motion  its  angular 
velocity  is  increasing  at  the  rate  of  5.81  velocity-units  per  sec- 
ond, i.e.,  5.81  radians  per  second  per  second. 

Another  method  is  this:  The  velocity  at  the  mid-point  of 
the  da  is,  as  before,  0.01745  -r-  0.01,  =  1.74  radians  per  sec- 
ond, while  at  the  middle  of  the  da'  it  is  0.01803  -f-  0.01,= 
1.803  radians  per  sec. ;  taking  the  difference  of  which  we  find 
that  0.058  rad.  per  sec.  of  angular  velocity  has  been  gained 
while  the  wheel  was  passing  between  these  two  mid-points. 
Hence,  dividing  this  gain  by  the  time  of  passage,  0.01  sec.,  we 
obtain  0  =  0.058  -r-  0.01  =  5.81  rad.  per  sec.  per  sec. 


APPENDIX. 


NOTES 

ON   THE 

GRAPHICAL  STATICS  OF  MECHANISM. 


PREPARED  BY  I.  P.  CHURCH,  CORNELL  UNIVERSITY. 

These  notes  are  based  mainly  on  the  work  of  Prof.  Herrmann  of  Aix-la- 
Chapelle,  the  earlier  form  of  which  was  presented  as  an  appendix  to  Vol. 
Ill,  of  Weisbach's  Mechanics. 

It  is  thought  that  the  form  of  presentation  adopted  in  the  following  pages 
is  that  best  adapted  for  students  already  familiar  with  the  Graphical  Statics  of 
quiescent  structures,  which  Prof.  Herrmann  assumes  is  not  the  case  with  the 
readers  of  his  book. 

For  greater  clearness,  all  force  polygons  have  been  placed  on  separate  parts 
of  the  paper  from  the  mechanism  itself,  instead  of  being  superposed  on  the 
latter  as  in  Prof.  Herrmann's  book. 

The  figures  referred  to  in  the  text  will  be  found  in  the  back  of  this  pam- 
phlet; while  the  paragraphs  (§)  referred  to  (of  a  higher  number  than  "30")  will 
be  found  in  the  writer's  "Mechanics  of  Engineering." 

For  table  of  contents  see  p.  28. 


1.  Assumptions.  The  forces  acting  on  each  part  of  any  mech- 
anism .here  treated  will  be  considered  to  be  in  the  same  plane 
and  to  form  a  balanced  system,  i.e.,  to  be  in  equilibrium ;  in 
other  words,  the  motions  of  the  pieces  take  place  without  sen- 
sible acceleration  (or  the  effect  of  inertia  is  disregarded).  (See 
p.  440  of  Prof.  Uri win's  Machine  Design  for  consideration  of 
the  inertia  of  a  piston.)  Also,  the  weights  of  the  pieces  will  be 
neglected  unless  specially  mentioned. 

*  2.  Graphical  Treatment.  A  "  two- force  "  piece,  or  a  two-force  " 
member  of  a  mechanism  is  one  acted  on  by  only  two  forces ; 
which  forces,  therefore,  must  be  equal  and  opposite  and  have  a 


2  NOTES   ON   THE    GRAPHICAL   STATICS   OF   MECHANISM. 

common  line  of  action,  for  equilibrium.  No  force  polygon 
need  be  drawn  for  such  a  piece. 

A  Three-Force  Piece.  Here,  for  equilibrium,  the  three  lines  of 
action  must  meet  in  a  common  point  and  the  force  polygon  is  a 
triangle  (see  §  325);  for  example,  see  Fig.  A  (bell-crank),  (p.  28). 

For  the  equilibrium  of  &  four-force  piece  (Fig.  B),  the  result- 
ant of  any  two  of  the  forces  must  be  equal  and  opposite  to  that 
of  the  other  two  forces.  The  common  line  of  action  of  these 
resultants  is  the  line  joining  the  intersections  a  and  &,  while  their 
common  amount  is  given  by  the  ray  00,  which  is  parallel  to 
a . .  b  and  is  a  diagonal  of  the  closed  force  polygon  (here  a  quad- 
rilateral). If  the  four  forces  act  in  parallel  lines,  and  two  are 
given,  we  determine  the  other  two  by  an  "equilibrium  poly- 
gon," etc.,  by  §  329 ;  i.e.,  we  treat  the  two  unknown  forces  as 
pier-reactions,  even  if  their  action-lines  (one  or  both)  are  between 
the  action-lines  of  the  known  forces.  (For  example,  see  [A]  in 
Fig.  IT,  Plate  Y,  where  from  the  known  forces  Sl  and  $,  we 
construct  the  two  unknown,  P  and  R,  in  given  action -lines,  to 
balance  them.  The  equilibrium  polygon  begins  at  c  in  the 
left-hand  abutment- vertical  and  consists  of  segments  c . . .  d, 
d. ..  e,  and  e .../*,  terminating  in  the  right-hand  abutment- verti- 
cal, at/*.  We  then  draw  c . .  .f  as  the  abutment-line,  or  "  closing 
line.") 

No  more  than  four  forces  will  act  on  any  piece. 

3.  Efficiency.  In  each  of  the  following  problems  there  will  be 
but  one  working  force  or  driving  force,  P,  and  but  one  useful 
resistance,  Q ;  all  other  resistances  being  due  to  friction.     For 
present  purposes  we  are  to  understand  by  efficiency  the  ratio  of 
the  value  P0,  which  would  be  sufficient  for  the  driving  force  if 
there   were  no  friction  of  any  kind,  in  order  to  overcome   Q 
(without  acceleration),  to  the  value  P,  which  it  must  actually 
have,  to  overcome  Q  under  actual  conditions,  i.e.,  with  friction. 

p 
Hence  efficiency  =  rj  =  -7^.     [Strictly,   the   efficiency    involves 

the  distances  traversed  ;  see  §  5.] 

4.  Backward  Motion.  If  the  useful  resistance  Q  is  a  load  due  to 
gravity,  i.e.,  a  weight,  the  value  to  which  the  working  force 


3  NOTES   ON   THE   GRAPHICAL   STATICS   OF  MECHANISM. 

(now  such  no  longer)  must  be  diminished  in  order  to  allow  the 
mechanism  to  run  backward  without  acceleration  will  be  called 
P  ';  and  if  in  any  case  P'  is  found  to  be  negative,  we  recognize 
the  mechanism  to  be  self-locking  •  that  is,  it  will  not  run  back- 
ward (or  "  overhaul  ")  when  the  working  force  is  zero,  but,  on 
the  contrary,  a  force  must  be  applied  in  the  action-line  of  the 
working  force  in  the  opposite  direction  to  cause  a  backward 
motion.  For  instance,  in  Fig.  C,*  the  force  P  is  necessary  for 
the  uniform  downward  motion  of  the  handle  A,  to  raise  Q,  and 
overcome  friction  at  all  points.  With  no  friction,  P0  would  be 

PQ 
sufficient  to  raise  Q,  and  the  efficiency  =  -77°;  whereas,  to  enable 

Q  to  sink  uniformly,  a  still  smaller  force,  viz.  P\  must  be 
applied  at  A  (but  still  positive  in  this  case,  so  that  the  machine 
is  not  self-locking). 

5.  Efficiency  in  General.  .  .  ,  with  one  working  force  P  and  one 
useful  resistance  Q,  is  the  ratio  of  the  work  Qs'  to  the  work  Ps, 
where  sr  and  s  are  the  respective  distances  worked  through  in 
forward  motion  by  Q  and  P  (simultaneously);  i.e., 

77,  or  efficiency,  =  -~  .......  (1) 


Now  by  §  142,  if  2(Fs")  denotes  the  sum  of  the  amounts  of 
work  lost  in  friction  at  various  points  where  rubbing  occurs  in 
the  mechanism,  we  have 

Ps  =  Qs'  +  2(Fs")  .......    (a) 

But  in  the  ideal  case  of  no  friction  (or  perfect  efficiency), 
taking  the  same  range  of  motion  as  before,  and  letting  P0  denote 
the  new  (and  smaller)  value  of  the  working  force  which  is  now 
sufficient  to  overcome  ,  we  have 


Hence  from  (#),  (#),  and  (1)  it  is  plain  that  the  efficiency  may 
be  written  thus  : 


the  form  proposed  in  §  3  above. 


*  On  p.  28  of  this  appendix. 


4  NOTES   OK   THE   GRAPHICAL   STATICS   OF   MECHANISM. 

6.  Condition  of  being  Self-Locking.  If  the  efficiency  is  less  than 
0.50  and  the  lost  work  of  friction  be  assumed  to  be  the  same  in 
forward  as  in  backward  motion   (in  same  range  of  motion,  of 
course),  the  mechanism  is  self-locking. 

Proof.  For  forward  motion  (no  acceleration),  letting  2(Fs") 
denote  the  sum  of  the  amounts  of  work  lost  in  friction  at  the 
various  points  of  rubbing,  we  have,  for  a  definite  range  of 
motion,  (see  §  142,) 

Ps  =  Qsf  +  2(1*8"), (2) 

and   for  backward   motion   (same   range),   similarly,  Q  being  a 
a  working  force  and  P'  a  resistance,  (see  §  142,) 

Qsf  =  P's  +  2(Fa") (3) 

As  implied  in  the  notation,  assume  that  the  friction-work  is 
the  same  in  backward  as  in  forward  motion  of  the  mechanism. 
From  (1)  and  (2)  we  deduce 

£l'  =  Qs'  +  2(Fs")  ;    or,  2(W)  =  Qs'  g  -  l].    .    .    (4) 
Substituting  from  (4)  in  (3)  we  obtain,  finally, 

P*  =  ^'[2.00  -  -1 (5) 

Si  T}_] 

From  (5)  it  is  evident  that,  when  the  efficiency  is  less  than 
0.50,  P'  is  negative ;  that  is,  the  mechanism  is  self-locking.  The 
assumption  made  above  as  to  equality  of  lost  work  in  forward 
and  backward  motion  is  not  exactly  true  for  any  machine,  per- 
haps. In  most  mechanisms  the  friction  work  is  somewhat  less  in 
backward  motion,  and  the  proposition  is  then  true  if  for  0.50  we 
put  a  smaller  value  for  r/. 

Machines  of  peculiar  design  may  be  constructed  with  rj  greater 
than  0.50,  which  nevertheless  are  self -locking ;  but  in  these  we 
find  that  the  lost  work  is  greater  in  backward  than  in  forward 
motion. 

7.  Sliding  Friction.    By  §  156  we  know  that  if  two  rough  sur- 
faces slide  on  each  other  the  mutual  action  or  force  between 
them  is  not  normal  to  the  plane  of  contact,  but  inclined  to  the 


5  NOTES   ON   THE   GRAPHICAL   STATICS   OF   MECHANISM. 

normal  at  an  angle  0,  the  "  angle  of  friction,"  on  that  side  of  the 
normal  opposed  to  the  direction  (of  relative  motion)  of  the  body 
under  consideration.  Thus,  Fig.  1,  Plate  I,  if  the  block  A  is 
sliding  toward  the  right  relatively  to  B  (no  matter  which  one,  if 
either,  is  absolutely  at  rest)  the  mutual  force  between  them  acts 
in  the  line  o  .  .  .  b  ;  if  toward  the  left,  then  it  acts  in  the  line 
o'  .  .  .  V.  The  pressure  of  B  upon  A  is  from  m  toward  o  (or  o')\ 
that  of  A  upon  B  from  m  toward  b  (or  &'). 

8.  Example!  Mill  Elevator.  (Plate  I,  Fig.  2.)  The  single 
rigid  body  Ac  consists  of  a  platform  and  vertical  side-piece, 
which  rubs  at  &,  the  left  side  of  the  fixed  vertical  guide  6y,  and 
also  at  #,  the  right  side  of  the  same. 

Let  Q  be  the  combined  weight  of  the  load  on  the  platform 
and  the  platform  itself,  and  P  the  required  pull  or  tension  to  be 
applied  vertically  at  A,  to  maintain  a  uniform  vertical  upward 
motion  (forward  motion,  here).  Besides  the  forces  P  and  Q, 
the  body  Ac,  considered  free,  is  acted  on  by  the  forces  J?x  and 
7?2  at  the  rubbing  surfaces  a  and  £,  acting  at  the  proper  incli- 
nation (0)  from  tneir  respective  normals  ;  note  which  side. 

Evidently  A  .  .  c  is  a  four-force  piece.  Four  action-lines  are 
known,  but  only  one  amount,  that  of  Q.  To  find  the  amounts 
of  P,  jftt,  and  R»  we  join  d  (intersection,  or  "co-point"  of  P 
and  J2,)  with  c  that  of  R^  and  Q.  On  the  right  of  the 
figure  we  begin  the  force-polygon  by  laying  off  m  .  .  o  parallel 
and  equal  to  Q  (by  scale).  After  drawing  through  m  a  line 
parallel  to  7?2,  and  through  o  a  line  parallel  to  c  .  .  d  (  the  known 
action-line  of  the  resultant  of  Q  and  7?2),  by  the  intersection  k 


we  determine  R^  =  in  .  .  k,  and  the  resultant,  o.  .  k,  of  Q  and  7?a. 
But  this  resultant  should  balance  P  and  7?,,  and  therefore  close 
a  triangle  with  them  in  the  force-polygon  ;  hence  parallels  to  P 
and  ^,  .  .  through  Ic  and  o  respectively,  finally  complete  the  force 
quadrilateral  m  .  .  n,  and  fix  the  values  of  P  and  R^  which  can 
now  be  scaled  off. 

In  this  figure,  for  clearness,  a  large  value,  about  20°,  has  been 
given  to  0,  so  that/*,  —  tan  0,=  about  0.36  ;  and  from  the  force- 
polygon  we  find  that  for  Q  =  110  Ibs.,  P  is  about  140  Ibs.  If 
there  were  no  friction  Rl  and  R^  would  be  horizontal,  and  a 


10 


6  NOTES   ON  THE   GRAPHICAL   STATICS   OF   MECHANISM. 

force  P0=Q  =  110  Ibs.  would  be  sufficient,  vertical  and  upward 
at  Ay  to  maintain  a  uniform  upward  motion  (or  to  permit  a  uni- 
form downward  motion,  once  started).  Hence  the  efficiency  for 

upward  (forward)  motion,  with  friction,  is  r)  =  -p  =  — -  =  0.78  or 

about  78  per  cent. 

In  uniform  backward  motion  with  friction,  Rl  and  7?8  (now 
call  them  R{  and  Rj )  will  change  their  action-lines  to  the  other 
sides  of  their  normals ;  i.e.,  they  will  now  act  along  a.  .d  and 
b  . .  c',  respectively,  at  an  angle  0  with  the  normals.  Hence  the 
line  G'  . .  d\  in  backward  motion,  takes  the  place  of  c . .  d,  in 
getting  the  new  force-polygon  (see  on  left  in  Fig.  2);  i.e.,  o'. .  k' 
is  drawn  parallel  to  G'.  .  d',  and  m'. .  kr  parallel  to  G'.  ,  b,  Q 
being  known.  Then  a  vertical  through  kf  and  a  parallel  to 
d'. .  a  through  o'  complete  the  polygon;  which  fixes  .P'as  well 
as  R{  and  RJ.  For  Q  =  110  Ibs.,  P'  is  roughly  about  80  Ibs. 

Since  P'  is  upward  (i.e.,  not  negative,  being  in  the  same 
direction  as  P)  the  machine  is  not  self-locking.  To  alter  the 
design,  however,  so  that  it  shall  be  self-locking,  we  need  only 
place  the  rubbing  point  a  near  enough  to  b  to  cause  a..d'to 
pass  through,  or  below,  G'\  for  then  o'. .  Tc!  will  either  coincide 
with  o' '. .  n'  or  pass  below  it,  thus  making  P'  either  0  (zero) 
or  negative.  But  .the  pressures  R^  and  R^  will  be  enormously 
increased.  Also,  P  for  upward  motion  will  be  larger,  and  the 
efficiency  smaller  than  before. 

9.  Example  II.  Wedge,  (Plate  I,  Fig.  3.)  Required  the  neces- 
sary horizontal  force  P,  at  the  head  of  the  wedge  AB,  to  raise  the 
load  Q  and  overcome  the  friction  at  the  three  points  of  rubbing, 
a,  c,  and  e.  For  the  upward  motion  of  the  block  E  (wedge  mov- 
ing to  the  right)  the  lines  of  pressure  at  these  points  of  rubbing 
are  a  ..b,  c.  .e,  and  c..d;  there  being  no  pressure  at  E.  The 
block  Ee  is  a  three-force  piece,  under  the  action  of  the  known  Q 
and  the  unknown  R^  and  Rl  (i.e.,  the  Rl  which  acts  from  G 
toward  d)  all  three  action-lines  being  given.  Hence  the  force- 
triangle  is  immediately  drawn,  viz.,  m  . .  n  . .  0,  and  the  amounts 
of  J??2  and  Rl  become  known  by  scale. 

The  wedge  AB  is  also  a  three- force  piece,  acted  on  by  the  Rl 


7  NOTES   ON   THE   GRAPHICAL   STATICS   OF   MECHANISM. 

(now  found)  pointing  from  c  toward  Z>,  and  the  unknown  7?3  and 
the  required  working-force  P.  Its  force  triangle  n  . .  o  . .  k  is 
then  drawn,  since  we  have  just  found  7?,  and  have  only  to  make 
o  . .  k  horizontal  ( i.e.,  parallel  to  P)  and  n  . .  ~k  parallel  to 
a  . .  5,  to  close  the  triangle  and  thus  determine  P  =  o  . .  k ;  (for- 
ward motion  with  friction). 

With  no  friction,  J?3  is  vertical,  7?2  horizontal,  and  Kl  follows 
the  normal  to  the  plane  A  . .  B ;  and  the  corresponding  force- 
triangles,  beginning  with  the  known  Q,  are  m0 .  .  n9 . .  o0  and 
n0 . .  o0 . .  k0 ;  and  thus  P0  is  found. 

From  the  drawing  we  have,  with  Q  =  102  Ibs.,  P  =  98  Ibs. 

30 
and  P0  =  about  30  Ibs.;  so  that  the  efficiency  i?  =  ^  ~  0.  31. 

i/O 

The  angle  0  used  is  about  20°  or/  =  0.  36. 

For  backward  motion,  7?/  would  act  parallel  to  c .  .  d', 
R^  parallel  to  e  . .  c' ',  and  l?sf  parallel  to  a . .  #';  and  the  corre- 
sponding force-triangles  are  m '. .  n'. .  o'  and  n'. .  o'. .  kf,  the  result- 
ing value  of  P'  being  negative.  That  is,  P'  must  act  from  right 
to  left,  for  backward  motion.  Hence  the  mechanism  is  self- 
locking  (for  these  particular  data,  in  Fig.  3). 

Note. — It  can  easily  be  shown  that  if  the  angle  of  the  wedge 
(i.e.,  the  angle  between  its  two  sides)  is  less  than  twice  the 
angle  of  friction  (supposed  the  same  for  all  three  rubbing 
contacts)  the  mechanism  is  self-locking.  (This  is  best  shown 
graphically.) 

10.  Example  III.  The  Jack-Screw.  ( Plate  I,  Fig.  4.)  Kequired 
the  value  of  each  force  P  of  a  couple,  in  a  horizontal  plane,  applied 
to  the  cross-bar  a0  J0,  in  order  to  raise  the  load  Q  and  overcome 
the  friction  between  the  surfaces  of  the  square-threaded  screw 
and  nut.  The  screw-shaft  is  vertical.  Assume  no  pressure  at 
the  edges  of  the  threads.  The  pressures  on  the  helical  surfaces 
may  be  considered  concentrated  at  two  points  dl  and  d^  diametri- 
cally opposite,  in  the  middle  of  the  width  of  the  thread.  The 
pressure  at  d^  viz.  J?,,  will  lie  in  a  vertical  plane  ~|  to  a0 . .  b0,  and 
make  an  angle  0  -f-  OL  (on  the  left)  with  the  vertical,  while  R^  on 
the  other  side  makes  an  equal  angle  with  the  vertical  (but  on  the 


NOTES   ON   THE    GRAPHICAL   STATICS   OF   MECHANISM. 

right),  a  is  the  angle  which  the  helix  of  screw  surface  makes 
with  the  horizontal. 

Projecting  the  five  forces  on  a  plane  ~|  to  the  bar  a-0 . .  &0,  we 
find  that  Q  must  balance  the  vertical  components  of  Rl  and  7?2, 
which  justifies  the  force-triangle  drawn  at  [#],  where,  with  Q 
given,  we  easily  fix  R^  and  7?2  by  drawing  w  . .  m  and  k. .  m  at 
an  angle  of  0  -f~  a  w^h  Qi  as  shown.  Therefore  the  horizontal 
component  of  J^l  isp  . .  m,  that  of  7?2  is  m .  .p. 

Projecting  all  the  five  forces  on  a  horizontal  plane,  we  note 
that  the  two  unknown  P's  must  balance  the  couple  formed  by 
the  horizontal  components  of  Rl  and  7?a,  as  if  these  components 
(which  we  will  now  call  Rz  and  7?4)  were  applied  directly  to  the 
bar  a0..b0  at  points  vertically  above  dt  and  d^  Hence  (see  \_C~\  in 
Fig.  4)  we  treat  the  known  Ez  and  7?4  (each  =  m .  .p)  as  par- 
allel forces  applied  perpendicularly  to  a  straight  beam  a . .  b 
supported  at  a  and  5  on  ||  smooth  surfaces  whose  normals  are  ~| 
to  a.  .l>  and  ||  to  Rz  and  7?4.  The  reactions  of  these  supports 
will  be  ||  to  7?3  and  7?4  and  are  the  forces  P  and  P  required.  At 
\_G]  in  Fig.  4,  dl . .  d^  =  the  distance  dt . .  d^  of  [yl],  and  a.  .b 
=  distance  a0 .  .  b0. 

Hence,  using  the  construction  of  §  329,  we  make  x . .  y  =  7?3, 


y . .  x  =  7?4,  select  a  pole  O  at  convenience,  draw  the  three  rays 
O..x,  O..y^  and  0..  x,  and  a  corresponding  equilibrium  poly- 
gon u  . .  t . .  s  . .  r,  beginning  at  any  point  u  in  the  action-line 
u . .  b  of  the  lower  P.  Join  u  . .  r,  the  abutment-line,  and  draw 
a  line  ||  to  it  through  the  pole  0  to  fix  n'\  then  x . .  n'  =  P  re- 
quired. 

With  no  friction,  0  =  0,  and  7?3  and  7?4  are  each  equal  to 
mQ .  .p,  instead  of  m..p;  and  P0  is  proportionally  smaller  than 
P.  In  this  figure  P0  is  about  two  thirds  of  P\  i.e.,  the  efficien- 
cy is  about  0.66. 

For  backward  motion,  R^  and  R^  must  act  on  the  other  sides 
of  their  respective  normals;  i.e.,  in  \_H\  we  should  use  a —  0 
instead  of  a  -f-  0,  and  if  in  that  case  a  were  less  than  0,  the 
point  m  would  fall  on  the  right  of  p,  and  P'  would  be  negative; 
i.e.,  the  screw  would  not  run  backward  (would  not  "  overhaul ") 
when  there  was  no  force  on  the  cross-bar ;  ( self-locking). 


NOTES   ON   THE    GKAPHICAL    STATICS    OF    MECHANISM. 

11.  Pivot  Friction.  (Plate  II,  Fig.  5.)  Since  the  frictions  at  the 
base  of  a  flat-ended  pivot  (see  §  168)  are  equivalent  to  a  couple  in 
which  eacli    force  is  \fR,  or  \fQ  with   present  notation,  and 
whose  arm  is  f  of  the  radius  r  (so  that  the  moment  of  the  couple 
is  f  fQ^t  we   may   suppose   the   pressure   concentrated  at  two 
opposite  points,  in  the  circumference  having  a  radius  =  f  r,  as< 
in  Fig.  5.     The  pressures  at  these  points  are  inclined  at  an  angle 
=  <p  with  their  respective  normals  and  in  the  proper  directions, 
as  shown  in  Fig.  5.     If  uniform  motion  is  to  be  maintained  by  a 
horizontal  couple  of  a  moment  =  Pa,  P  being  unknown  and  Q 
and  0  given,  we  may  proceed  graphically  as  in   the  preceding 
problem,  making  a  =  0  (zero);  a  here  denotes  the  distance  a  . .  b 
of  preceding  problem. 

12.  Journal  Friction.    Assuming  that  the  journal  is  fitted  to  its 
bearing  with  sufficient  play  to  permit  the  line  of  contact  to  take 
any  position  (in  the  circumference  of  the  bearing)  called  for  by 
the  conditions  of  equilibrium  (i.e.,  there  is  no  initial  clamping), 
we  know  from  §163  that  the  action-line  of  the  mutual  pressure 
between  them  must  be  tangent  to  the  friction-circle,  when  rub- 
bing is  taking  place.     As  to  which  side  of  the  friction-circle  is  to 
have  the  action-line  drawn  tangent  to  it,  that  is  a  matter  depend- 
ing on  the  direction  of  the  motion  and   must  be  decided  by  the 
statement  in  §  7.    The  bearing  in  which  the  journal  turns  may 
itself   be   in    motion  (the   crank   end   of   a   connecting-rod,  for 
instance)  and  the  direction  of  relative  turning  must  be  considered. 
The  following  examples  will  bring  out  clearly  all  the  relations  in 
such  cases.     Each  friction-circle  will  be  much  exaggerated,  for 
clearness,  in  the  figures,  and  must  not   be  mistaken  for  a  journal. 
The  radius  of  a  friction  circle  is  r  sin  0,  where  r  is  the  radius  of 
the  journal.     (Note  carefully  the  relations  in  §  163.) 

is.  Example  IV.  Bell-crank.  (Plate  II,  Fig.  6.)  The  load  Q  is 
given  in  amount  and  direction,  being  supported  by  a  vertical 
link  which  is  jointed  at  A  to  the  bell-crank.  The  working  force, 
PI  has  a  given  direction  and  is  applied  through  a  link  jointed  at 
B  to  the  bell-crank.  Both  joints  consist  of  journals  in  bearings. 
P  —  ?  to  overcome  friction  at  A,  B,  and  (7,  and  to  raise  Q.  The 
motions  at  A.  and  B  are  such  that  the  action-line  of  Q  must  be 


10  NOTES   ON   THE   GRAPHICAL   STATICS   OF  MECHANISM. 

tangent  to  A' a  friction-circle  on  its  left;  and  that  of  P  tangent 
to  B?$>  friction-circle  on  its  under  side. 

Drawing,  then,  these  tangents  ||  to  the  respective  known  di- 
rections of  P  and  Q,  we  have  A  . .  a  and  c . .  a  as  their  action-lines, 
meeting  at  #,  through  which  (since  the  bell-crank  is  a  three-force 
piece)  the  third  force  R  must  pass  (the  reaction  at  (7),  and  this 
must  be  tangent  to  the  friction-circle  at  C  on  its  upper  side.' 
Hence  a  line  through  a  and  tangent  to  the  friction-circle  at  C  on 
upper  side,  is  the  action  line  of  It,  viz.,  a . .  b.  The  force-trian- 
gle ~k . .  m . .  n  is  now  easily  drawn,  Q  being  the  known  force  and  laid 
off  first,  and  P  =  m . .  n  is  thus  determined.  With  no  friction,  P, 
arid  Q  (directions  unchanged)  must  pass  through  the  centres  of 
the  pins  at  A  and  B,  and  intersect  at  a0.  R  then  passes  through 
a0  and  the  centre  of  C,  i.e.,  acts  along  ajbQ.  P0  =  mnQ  in  the 
new  force-triangle,  and  the  efficiency  =  m..n0  -=-  m..n.  For 
backward  motion  we  make  the  action-lines  tangent  to  their  re- 
spective friction-circles  on  the  other  side  in  each  case ;  then  P' 
=  m.  .n'. 

14,  Example  V.  The  Slider-crank.  (Plate  II,  Fig.  Y.)  The 
wheel  TFand  the  crank  B  form  a  single  rigid  body  turning  on  a 
fixed  bearing  C.  The  connecting-rod  or  link,  BD,  is  pivoted 
about  the  crank-pin  at  one  end  and  to  the  cross-head  pin  at  the 
other.  The  cross-head  block  slides  in  a  right  line  between  guides, 
and  receives  the  pull  (or  push)  of  the  piston-rod  in  the  axial  line 
of  that  rod,  i.e.,  through  the  centre  of  the  pin  at  D.  The  resist- 
ance Q  being  given,  applied  in  line  a..~b  to  wheel  W,  required 
the  necessary  value  of  P  for  uniform  motion  m  ike  given  position 
of  the  mechanism.  The  connecting-rod  is  a  two-force  piece,  and 
in  its  present  position  (B  on  the  right  of  a  vertical  through  C) 
the  pressure  at  the  cross-head  pin  is  tangent  to  the  friction-circle 
there  on  its  upper  side  ;  that  at  the  crank-pin  on  the  lower  side 
of  the  friction -circle  there.  Hence  a  line  drawn  so  as  to  be  tan- 
gent to  the  two  circles  in  the  manner  stated  is  the  action-line  of 
R^  the  crank-pin  pressure  (as  also  that  at  the  cross-head  pin);  i.e., 
draw  e . .  h. 

The  three-force  piece,  WCB^  is  acted  on  by  the  known  Q,  by 
RV  and  a  pressure  at  the  bearing  (7,  viz.,  7?2,  which  is  tangent  to 


11  NOTES   ON   THE    GRAPHICAL   STATICS   OF   MECHANISM. 

the  friction-circle  at  C  on  the  upper  side ;  therefore  7?2  must  act 
in  the  line  b .  ./"drawn  from  the  intersection  b,  of  Q  and  Rl  tan- 
gent to  friction-circle  at  C  on  upper  side.  Hence  the  force-tri- 
angle for  WCB  is  easily  completed  by  making  o..m||and  = 
Q,  and  drawing  m . .  n  and  o . .  n  \\  to  c . .  h  and  f. .  b,  respectively, 
thus  determining  Itl  and  R^.  Now  the  cross-head  block,  Dk,  is 
a  three-force  piece  acted  on  by  Rl  (pointing  toward  the  left),  now 
known  ;  by  the  unknown  P  and  by  the  unknown  R,  which  is  the 
pressure  coming  from  the  upper  guide,  and  must  act  in  a  line  d . .  e 
through  the  intersection  of  the  action-lines  of  P  and  R^  viz.,  e, 
and  makes  angle  =  0  (in  direction  as  shown)  with  the  normal  to 
the  guide  surface.  Its  force-triangle,  then,  is  rks,  k .  .  r  being  = 
and  opposite  to  m ..  n,  s..r  being  drawn  ||  to  P  and  k..s\\to 
d.  .e.  Thus  P  is  found,  being  =  rs. 

As  for  P0  (i.e.,  with  no  friction),  R^  would  act  through  the 
pin  centres,  thus  raising  b  ;  Rz  would  act  through  ~b  and  the  centre 
of  C\  while  J?3  would  be  normal  to  the  guide  surface.  With  new 
force-triangles,  on  this  basis,  we  find  P0  =  r . .  s0;  whence  the  ef- 
ficiency, =  P0  -f-  P,  is  found. 

In  any  other  position  of  the  mechanism  a  similar  method  is 
available. 

15,  Example  VI.  Beam-engine  with  Evans's  straight-line  motion. 
(Plate  II,  Fig.  8.)  WKM  is  a  single  rigid  body  (wheel  and 
shaft)  turning  on  a  fixed  bearing  at  K.  M  is  the  crank-pin,  MB 
the  connecting-rod,  DC  &  link  turning  in  a  fixed  bearing  at  D, 
and  pivoted  at  C  to  the  beam  ACE,  one  end  of  which,  E,  is 
guided  in  a  horizontal  right  line  by  the  block  7^  and  straight 
guides.  The  pin  C  being  mid-way  (in  a  straight  line)  between  E 
and  A,  and  the  length  CD  being  made  equal  to  AC,  which  also 
=  CE  (between  centres;  D  and  E  at  the  same  level),  A  will  be 
compelled  to  move  in  a  vertical  straight  line,  as  if  it  were  itself 
guided  by  a  straight  edge.  The  vertical  piston-rod  is  linked  to 
the  beam  at  A.  Required  the  necessary  steam  pressure  P  on  the 
piston-head,  for  upward  -motion,  the  resistance  Q  being  applied 
to  the  wheel  W  in  the  line  x . .  w,  and  the  parts  having  the 
position  shown  in  the  figure. 

The  link  DC  is  a  two-force  piece,  subjected  at  this  instant  to 


12  NOTES   ON    THE    GRAPHICAL    STATICS    OF   MECHANISM. 

some  thrust,  R^  whose  action-line  must  be  tangent  (below)  to  the 
friction-circle  at  C,  and  also  (above)  to  that  at  D.  Similarly  the 
link  or  connecting-rod  BM  is  a  two-force  piece,  under  a  tension 
7?,,  whose  action-line  is  tangent  to  the  friction-circle  both  at  M 
and  B.  At  M.  thistangency  is  on  the  right ;  at  B  it  may  be  either 
on  the  right  or  the  left  according  as  the  link  has  not  yet 
reached,  or  has  passed,  a  certain  critical  position  which  is  very 
nearly  the  position  chosen  for  the  figure ;  in  which  the  tan- 
gency  has  been  drawn  on  the  left  at  B. 

The  block  J^is  a  two-force  piece,  under  a  thrust  7?4,  directed, 
as  shown,  at  an  angle  0  with  the  normal  to  the  guide-surface 
below  it. 

Construction.  We  first  draw  the  force-polygon  for  TPjOf,  a 
three-force  piece,  the  forces  being  the  known  Q  in  line  w.  .a?,  the 
unknown  JSl  in  the  known  line  c. .  A,  and  the  unknown  reaction 
R^  at  the  bearing  K. 

Rz  must  act  in  a  line  x  . .  y  through  the  point  a?,  and  tangent, 
on  left,  to  the  friction-circle  at  I£.  Hence  by  laying  off  o  . .  m  = 
and  ||  to  Q,  and  then  drawing  m . .  n  \\  to  h . .  c  and  o  .  .  n  \\  to 
x . .  y,  we  determine  Rl  and  J?2. 

The  beam  ABCE\§  a  four-force  piece  under  the  forces  7?1?  7?3, 
7?4,  and  RV  their  action-lines  being  all  known  and  Rl  now  known 
in  amount,  =  m  .  .n.  The  direction  of  JR3  becomes  known  from 
a  consideration  of  the  piston  and  rod  which  together  form  a 
three-force  piece,  being  acted  on  by  jP,  by  the  stuffing-box  pres- 
sure 7?6,  and  by  7?3  reversed.  Since  P  and  R6  intersect  at  #,  and 
since  JK3  must  pass  through  a  and  be  tangent  on  the  right  to  the 
friction-circle  at  A,  its  action-line  is  easily  drawn.  Resuming  the 
body  ABCE,  we  find  the  intersection,  e,  of  Rz  and  Rl  (N.B.  e  is 
off  the  paper)  and  join  e  with/*,  the  intersection  of  7?4  and  7?6. 
Making  n1..ml  =  and  ||  to  R^  (now  pointing  down),  and  draw- 
ing nl . .  r  ||  to  f . .  e  and  m, . .  r  ||  to  R3 ;  also  r  . .  u  \\  to  7?4  and 
ft,, . .  u  ||  to  RS  :  we  finally  determine  7?3,  R^  and  R^ 

Having  found  7?3,  the  force-triangle  for  the  three-force  piece 
A  . .  a  (piston,  etc.)  is  easily  drawn  (see  [Z]  in  Fig.  8)  and  P  finally 
determined. 

With  no  friction  we  find  P0  by  taking  friction-circle  centres 


13  NOTES   ON   THE   GRAPHICAL   STATICS   OF   MECHANISM. 

instead  of  tangencies  and  making  R6  and  Rt  ~]  to  their  respective 
sliding  surfaces. 

16.  Example  VII.  Oscillating  Engine.  (Plate  III,  Fig.  9.) 
Here  the  connecting-rod  is  dispensed  with,  the  piston  acting 
directly  on  the  crank-pin  A,  while  the  cylinder  EE  oscillates  on 
trunnions  turning  in  fixed  bearings.  The  crank  AB  and  wheel 
W constitute  a  single  rigid  body  turning  in  a  fixed  bearing  B. 
The  resistance  Q  being  applied  to  W  in  line  d . .  c,  we  wish  to 
find  the  proper  steam  pressure  P  on  the  left  of  the  piston  to 
overcome  Q  and  all  intervening  frictions,  for  motion  (with  in- 
sensible acceleration),  of  the  mechanism  in  the  position  shown.  P 
acts  centrally  along  the  axis  of  the  piston-rod.  The  pressure  R± 
is  that  of  the  stuffing-box  against  the  piston-rod,  and  Rz  that  of 
the  cylinder  against  the  edge  of  the  piston.  Remembering  that 
the  piston  and  cylinder  always  have  a  common  axis  as  they  "  tele- 
scope" in  and  out,  we  see  that  there  are  only  two  forces  external 
to  these  two  pieces  when  considered  together,  viz.,  the  pressures 
at  the  crank-pin  and  at  the  trunnion,  which  two  pressures  (J?2) 
must  therefore  be  equal  and  opposite,  both  of  them  acting  in  the 
line  h . .  g  tangent  to  the  two  friction-circles  (on  the  lower  side  at 
both  A  and  0).  We  have  thus  found  the  action-line  of  7?2. 

Now  consider  the  three-force  piece  WBA.  The  forces  are 
the  known  Q  in  line  d. .  c9  the  unknown  7?2  in  the  line  h . .  g,  and 
the  reaction  or  pressure  R \  in  a  line  a  . .  e  which  we  draw  through 
0,  the  intersection  of  R  and  (),  and  make  tangent  (on  the  upper 
side)  to  the  friction-circle  at  the  bearing  B.  Hence,  making  m  . .  n 
=  and  i|  to  Q,  and  m . .  r  and  n . .  r  ||  to  e . .  h  and  e..a9  respect- 
ively, we  fix  the  amounts  of  R^  and  R^  by  this  force-triangle 
m  . .  n . .  r.  * 

The  piston  is  a  four-force  piece,  under  the  action  of  7?2,  now 
known,  acting  toward  the  left  in  line  h . .  e ;  by  the  unknown  P 
acting  along  the  centre  line  of  the  piston-rod ;  and  by  the  two 
unknown  reactions,  Rz  and  R^  inclined  at  angle  (p  to  their  re- 
spective normals,  and  acting  at  known  points.  Hence  join  y, 
the  intersection  of  7?2  and  JR3,  with  a?,  that  of  P  and  R^  make  o , .  p 
=  and  ||  to  RV  draw^? . .  s  \\  to  R^  and  o . .  s  \\  to  y . .  a?,  to  deter- 


IT 


14  NOTES   ON   THE   GRAPHICAL   STATICS   OF   MECHANISM. 

mine  s ;  s . .  t  \\  to  P  and  o . .  t  \\  to  R^  are  then  drawn  to  fix  t.   P 
can  now  be  scaled  off,  as  also  Rz  and  R^. 

With  no  friction,  R^  would  act  in  the  axis  of  the  piston 
through  the  centres  of  A  and  C.  In  place  of  e  we  would  have  e0 
(not  shown  in  figure),  and  R  would  act  through  e0  and  the  centre 
of  B.  J?3  and  R±  would  be  ~]  to  their  respective  rubbing  surfaces. 
We  would  then  obtain  R^  =  m  . . r0  and  P0  =  Rz  =  m. .  r0 ; 
whence  the  efficiency  is  =  P0  -r-  P  =  m . .  r0  -f-  s . ,  t. 

16.  Example  VII.  The  Blake  Ore-crusher.  (Plate  III,  Fig.  10.) 
JTand  L  are  fixed  walls,  HD  oscillates  about  a  fixed  bearing  _Z>1? 
WFa  is  a  wheel  and  crank  rotating  on  a  journal  in  a  fixed  bear- 
ing F.  The  connecting-rod  a..c  communicates  motion  to  the 
two  links  or  struts,  DE  and  CB^  forming  a  toggle-joint  and 
causing  IfD1  to  oscillate. 

For  the  given  position  of  the  parts,  Q  being  the  resistance 
offered  by  a  piece  of  ore,  A,  to  any  further  motion  of  H  toward 
the  left,  required  the  value  of  P,  applied  in  the  line  W. .  b  to 
wheel  WF  to  overcome  Q  and  the  pressures  at  the  seven  sockets 
or  bearings. 

The  two  links  Z>J^and  CB  are  evidently  two-force  pieces,  DE 
being  subjected  to  some  thrust  Rz  in  line  h . .  c,  CB  to  some 
thrust  Rz  in  line  c . .  B  ;  these  lines  being  drawn  tangent  to  the 
various  friction-circles  in  the  manner  shown. 

Rz  and  Q  intersect  in  A,  hence  the  line  f . .  A,  drawn  through 
h  and  tangent  to  the  friction-circle  at  Z\,  must  be  the  action-line 
of  RV  the  reaction  at  the  bearing  Dr  That  is,  Q,  R^  and  7?2, 
are  the  forces  acting  on  the  three-force  piece  IIJ)l ;  therefore, 
Q  and  the  three  action-lines  being  known,  we  easily  complete 
the  corresponding  force-triangle  o . .  m  . .  n  and  thus  determine 
R,  and  Rz. 

Passing  to  the  three-force  piece  a . .  b  we  have  the  action-lines 
of  7?2  and  7?3  already  intersecting  at  c,  and  the  amount  of  7?2  = 
o  .  .  n.  The  third  force  7?4  has  an  action-line  passing  through  c 
and  tangent  (on  left)  to  the  friction-circle  at  the  crank-pin. 
Hence  draw  c. .  a  accordingly,  and  complete  the  force-triangle  by 
making  ~k  . .  i  \\  and  =  to  o  . .  n,  (i.e.,  to  7?2),  and  i . .  r  and 


15  NOTES   ON  THE   GRAPHICAL  STATICS  OF  MECHANISM. 

Jc.  .r  ||  to  a. .  c  and  B . .  c  respectively ;  which  determines  R^ 
and  J?4. 

Finally,  the  three-force  piece  a  . .  F. .  TFis  seen  to  be  acted  on 
by  RV  now  known  ;  by  the  unknown  P  in  line  P . .  &  ;  and  by 
the  bearing-reaction  R  acting  through  b  and  tangent  (on  right)  to 
the  friction-circle  at  F.  Draw  b . .  F^  then,  as  indicated,  and  the 
force-triangle  is  readily  formed,  t . .  s . .  u,  in  an  obvious  manner, 
whence  s . .  u,  =  P,  is  scaled. 

Without  friction  we  would  as  usual  draw  the  action-lines  of 
the  divers  jft's  through  the  centres  of  the  various  journals,  or 
sockets,  instead  of  tangent  to  their  friction-circles.  With  a  new 
set  of  force-triangles  on  this  basis  (see  broken  lines  on  the  right 
in  Fig.  10),  we  obtain  jP0,  considerably  smaller  than  P.  In  this 
exaggerated  figure  the  efficiency  is  only  about  0.50 ;  but  in  Prof. 
Hermann's  drawing,  withy7  for  journal  friction  =  0.10,  he  ob- 
tains a  value  of  0.80  for  the  efficiency. 

17.  Rolling  Friction  (so  called).  This  kind  of  resistance  is  due 
to  the  fact  that  the  point  of  application  of  the  force  acting  be- 
tween a  wheel  and  the  rail  or  surface  on  which  it  rolls  is  not  at 
the  foot  of  the  perpendicular  dropped  from  the  centre  of  the 
wheel  upon  the  rail,  but  a  little  in  front  (in  direction  of  rolling) 
by  an  amount,  or  distance,  ~b  =  about  0.02  in.  for  iron  wheels  on 
an  iron  rail,  and  about  0.50  in.  for  a  wagon-wheel  on  a  dry 
macadamized  road  (b  =  from  2."00  to  3/'00  on  soft  ground). 
See  §172;  (also  Prof.  Reynoids's  article  in  the  Philos.  Transac., 
vol.  166.)  As  to  the  direction  of  the  pressure  of  rail  on 
wheel  it  is  somewhere  within  the  cone  of  friction  so  long  as 
perfect  rolling  (i.e.,  no  slipping)  proceeds,  being  the  equal 
and  opposite  of  the  resultant  of  all  the  other  forces  acting  on  the 
wheel. 

Thus,  Fig.  11,  Plate  III,  if  Q  is  the  weight  of  the  roller, 
applied  in  the  centre  o',  the  necessary  force  P,  horizontal  and 
acting  through  the  centre  of  the  roller  (to  maintain  a  uniform 
rolling  motion),  is  determined  by  the  fact  that  the  resultant  of  P 
and  Q  must  act  through  c,  and  therefore  along  the  line  o' '. .  c, 
where  c . .  d  =  the  distance  1)  just  mentioned.  Hence,  making 
the  line  Q  through  o  equal  by  scale  in  length  and  ||  to  Q  at  o', 


16  NOTES   ON   THE   GRAPHICAL   STATICS   OF   MECHANISM. 

and  drawing  o  . .  m  \\  to  o' . .  c,  as  well  as  a  horizontal  through  the 
lower  extremity  of  Q,  we  determine  both  P  and  the  rail  pressure. 

Again,  Fig.  12,  if  the  rolling  action  occurs  on  both  sides  of 
the  roller,  as  at  c  and  e  when  a  weighted  plank  is  moved  horizon- 
tally  (to  left,  here)  on  loose  rollers,  we  note  that  it  is  a  two-force 
piece  (neglecting  the  weight  of  the  roller),  the  action-line  of  the 
compressive  forces  being  e . .  c,  both  e  and  c  having  been  located 
at  distance  —  b  from  the  perpendicular  through  the  centre  (in 
the  proper  directions ;  and  then,  we  may  have  different  Z»'s  at 
the  two  points  of  rolling  contact).  Here  the  upper  plank  is 
moving  and  the  horizontal  force  (toward  left)  which  must  be 
applied  to  it  to  maintain  this  motion  uniformly  must  be  equal  to 
the  sum  of  the  horizontal  components  of  the  various  inclined  72's, 
one  from  each  roller,  the  sum  of  the  vertical  components  of  the 
latter  being  equal  to  the  total  load  on  the  plank. 

In  the  uniform  motion  of  a  car-wheel,  EA,  Fig.  13  (brakes 
not  on),  the  (double)  wheel  and  its  axle  constitute  a  single,  rigid, 
two-force  piece,  acted  on  by  the  rail  pressure,  or  reaction,  at  c 
(c  . .  d  being  made  —  5)  and  by  the  pressure  of  the  bearing  against 
the  journal  or  axle  at  R\  and  this  pressure  must  be  tangent  to 
the  friction-circle  at  I?  (ou  the  right  for  motion  here  shown,  car 
moving  to  the  left).  The  horizontal  component  of  the  equal  and 
opposite  of  this  R  is  the  tractive  resistance,  on  a  straight  level 
track  with  uniform  motion  (besides  the  resistance  of  the  air),  and 
is  continually  overcome  by  the  tension  in  the  draw-bar  of  the 
locomotive.  R  is  practically  equal  to  its  own  vertical  component 
which  equals  the  portion  0  of  the  car's  weight  borne  by  this 
wheel,  and  this  horizontal  component  of  R  is  =  G  tan  6,  where 
6  denotes  the  complement  of  the  angle  e . .  c  . .  d. 

If  the  brake  is  in  action,  Fig.  D*  exerting  a  pressure  P'  at  the 
point  0,  this  pressure  must  act  along  the  line  a  . .  o  at  an  angle  = 
<pf  with  the  radius  aE  (note  the  direction  of  motion,  etc.),  pro- 
vided the  wheel  is  not  held  fast  by  the  brake.  Suppose  now 
that  the  wheel,  while  continuing  to  roll  without  slipping  on  the 
rail  is  on  the  point  of  slipping  upon  it,  that  is,  suppose  that "  skid- 
ding "  is  impending  ;  then  P' ',  the  pressure  of  the  rail  on  the 

*  See  p.  28. 


17  JJOTES   ON"   THE    GRAPHICAL   STATICS   OF   MECHANISM. 

wheel,  besides  being  applied  at  c,  must  take  the  direction  c . .  0, 
at  an  angle  <t>"  with  the  vertical  (track  level),  on  the  right,  here. 
Since  P'  and  P"  intersect  at  o,  the  only  remaining  force  acting 
on  the  rigid  body  JEW  (which  now  is  a  three-force  piece)  must 
pass  through  o.  This  force  is  P" ',  the  pressure  of  the  bearing 
on  the  journal,  and  it  must  also  be  tangent  to  the  friction-circle, 
at  E.  Its  action-line,  therefore,  is  easily  drawn  and  is  e  . .  o.  A 
corresponding  force-triangle  being  now  constructed  (not  shown  in 
the  figure)  with  sides  ||  to  0  . .  0,  a . .  0,  and  e . .  o,  respectively,  in 
which  the  vertical  projection  of  P"  is  made  —  6r,  the  portion  of 
weight  of  car  coming  on  this  wheel  (double),  we  determine  the 
value  of  each  force.  Since  <p"  for  impending  slip  (friction  of 
rest)  is  greater  than  0'  for  actual  slipping,  greater  resistance  is 
offered  by  impending  than  by  actual  "  skidding."  In  most  cases 
in  practice  the  distance  c . .  d,  =  0,  is  so  small  that  its  effect  in 
graphical  work  is  almost  inappreciable. 

18.  Example  VIII.  Friction-rollers  of  Crane.  (Plate  III,  Fig.  14.) 
At  T. .  X. .  Y  we  have  a  vertical  projection  of  the  crane.  The 
horizontal  pressure  of  the  crane  at  the  base  is  exerted  through 
friction-rollers  (shown  at  D  and  B)  against  the  side  of  the  fixed, 
conical,  and  vertical  mast  e  . .  0.  The  vertical  pressure  induced 
by  the  suspended  load  being  supported  at  T7,  while  a  horizontal 
pressure  simply  (  =  Q)  is  produced  at  the  base  X . .  F,  we  shall 
consider  the  equilibrium  of  the  part  G- . .  E,  carrying  the  friction- 
rollers,  as  if  it  were  acted  on  by  the  force  $,  applied  centrally 
and  symmetrically  in  the  line  a . .  Q . .  ~b  ;  by  the  pressures  /?, 
and  7?2  of  the  friction-roller  journals  against  their  bearings  ;  and 
by  a  force  P,  applied  on  the  lug  a,  and  ||  to  face  GE.  This 
force  P  is  of  such  an  amount,  to  l}e  determined,  as  to  maintain  a 
uniform  motion  in  direction  of  the  dotted  arrow. 

That  is,  EFGII  is  to  be  treated  as  a  four-force  piece.  Now 
each  friction-roller  is  a  two-force  piece  like  the  car-wheel  in  Fig. 
13  and  the  action-line  of  the  compressive  forces  in  each  is  drawn 
in  the  same  manner  as  in  Fig.  13,  noting  well  the  direction  of 
motion.  "We  thus  determine  the  action-lines,  f..  o  and  d..e, 
meeting  at  0,  of  the  unknown  Rl  and  7?3. 

Note. — Of  course  the  force  7?,  with  which  the  bearing  of  our 


18  NOTES   OX   THE   GRAPHICAL   STATICS   OF   MECHANISM. 

four-force  piece  acts  against  the  journal  of  the  friction-wheel  is 
the  equal  and  opposite  of  the  El  with  which  that  journal  acts 
against  the  bearing.  The  bearing  is  part  of  our  four-force  piece, 
and  the  force  R^  acting  upon  it  is  not  shown  in  the  figure,  but 
has  the  same  action-line  as  the  Hl  shown. 

Having  now  four  action-lines  and  the  amount  of  one  force, 
viz.,  Q,  of  the  four  forces  acting  on  the  piece  EFGH,  we  proceed 
to  construct  the  amounts  of  the  other  three,  in  the  way  so  often 
employed,  classifying  the  set  of  forces  into  two  pairs.  Pair  off 
P  and  Q ;  they  meet  at  a ;  join  a  . .  o  ;  lay  off  ~o" . .  in  =  and  || 
to  Q  ;  draw  m  . .  a"  \\  to  P  and  o" . .  a"  \\  to  o  ..a  thus  fixing 
P,  =  m  . .  a."  Also  draw  o".  .n\\to  d  . .  e  and  a" . .  n  \\  to 
o. ./,  thus  determining  the  amounts  of  R^  and  Rr 

Since,  with  no  friction,  P0  —  0,  the  efficiency  =  0,  i.e.,  all 
the.  work  done  by  P  is  wasted  (spent  in  overcoming  friction). 

19.  Chain  Friction.  (Plate  IV,  Fig.  15.)  Supposing  that  the 
groove  in  the  periphery  of  the  pulley  A  is  properly  constructed  for 
receiving  a  chain  in  such  a  way  that,  as  the  chain  winds  upon  the 
pulley,  the  alternate  links  place  themselves  with  their  planes  j| 
to  or  "I  to  the  axis  of  the  pulley  ;  then,  as  each  link,  c,  approaches 
the  point  K  where  it  is  to  assume  its  proper  place  on  the  circum- 
ference, its  neighbor  just  above  being  already  in  place  on  the 
pulley,  and  turning  with  it  (whereas  the  first  one  remains  vertical 
•for  a  time)  a  turning  of  one  in  the  hollow  of  the  other,  with 
corresponding  friction,  is  brought  about,  and  in  such  a  direction 
that  the  pressure  between  the  two  is  tangent  to  the  friction-circle 
on  that  side  of  the  latter  which  is  further  from  the  centre  of  the 
pulley  ;  for  one  turns  in  the  other  like  a  journal  in  the  bearing. 
(The  pulley  in  the  figure  is  turning  clock-wise.) 

Similarly,  on  the  other  side,  at  Z,  where  the  chain  is  unwind- 
ing and  thus  being  straightened,  each  link  turns  in  the  hollow  of 
its  neighbor,  on  passing  off  the  pulley,  and  hence  the  pressure  at 
L  is  tangent  to  the  friction -circle  there  on  the  side  nearer  to  the 
pulley-axle.  Again,  at  the  bearing  of  the  journal  of  the  pulley- 
axle,  the  reaction,  R,  is  tangent  to  the  friction-circle  and  on  its 
right. 

Let  r'  denote  the  radius  of  the  hollow  at  the  extremity  of  a 


19  NOTES   ON  THE   GEAPHICAL   STATICS  OF  MECHANISM. 

link  (equal  to  the  half-thickness  of  the  chain-wire)  and  r"  the 
radius  of  the  journal  of  the  pulley-axle;  also  0'  and  0"  the 
respective  friction-angles  at  those  points.  Then  the  radius  of 
the  friction-circle  at  L  and  at  K  =  a  =  r'  sin  0',  and  that  at  the 
central  bearing  —  b  =  r"  sin  <p"  '.  Let  r  denote  the  distance 
between  the  centre  at  K  (or  L)  arid  that  at  the  middle  bearing. 

We  thus  see  that  a  weight  Q  supported  on  the  ascending  side 
of  the  chain  (on  the  left,  here)  has  a  lever-arm  =  r  -|-  (a  -f-  b) 
about  the  point  of  application  of  R  ;  while  the  vertical  force  P, 
applied  to  the  descending  (and  unwinding)  side  of  the  chain,  has 
a  lever-arm  of  only  r  —  (a  -f-  £).  Hence  for  uniform  motion  we 
have  (from  equality  of  movements) 

P  :  Q  ::  r  +  (a  +  1)  :  r  —  (a  +  I). 

From  this  proportion  P  may  be  computed  ;  or,  graphically, 
given  Q  as  a  load  applied  to  a  horizontal  beam  with  supports 
at  R  and  P,  we  may  construct  R  and  P  as  if  reactions  of  those 


supports,  by  §  329.  Thus,  at  [^]  in  Fig.  15  make  m..R  \\  and 
=  Q;  take  any  pole  0  and  any  point  e  in  the  vertical  action-line 
of  R.  Draw  the  rays  0  .  .  m,  and  0  .  .  R  and  the  segments 
e..c,  and  c.  .n  \\  to  them  (respectively)  ;  join  e.  .n  and  make 
0  .  .  ri  ||  to  it  through  O.  Then  R  .  .  n'  =  P,  and  n'.  .  m  =  R. 

With  no  friction  P0  —  Q,  and  we  can  find  the  efficiency. 

For  backward  motion  the  lines  of  action  of  the  three  forces 
have  their  friction-circle  tangencies  each  on  the  side  opposite  to 
that  in  forward  motion,  and  we  have 

P'  :  Q::r-(a  +  b):r  +(a  +  l). 

20.  Example  IX.  Pulley  Blocks  or  Tackle.  (Plate  IY,  Fig.  16.) 
Let  C  be  the  upper  block  of  a  tackle  of  two  blocks  with  three 
pulleys  in  each  (the  pulleys  turning  independently  on  a  common 
bolt).  C  is  suspended  from  a  support  and  is  stationary,  and  R 
is  the  tension  in  the  rod  or  shank  by  which  it  hangs. 

The  lower  block  is  movable  and  carries  a  load  Q,  whose  uni- 
form upward  motion  is  to  be  maintained  by  the  application  of  a 
proper  force  P  or  #„  to  the  chain  at  the  extremity  ~b.  The  chain 
passes  continuously  over  the  six  pulleys  or  sheaves  in  the  manner 
shown  in  the  figure,  the  other  extremity  being  attached  to  the 


20  NOTES   ON   THE   GRAPHICAL   STATICS   OF   MECHANISM. 

support  above.  Each  straight  part  of  the  chain  will  be  con- 
sidered vertical,  and  all  these  straight  parts  have  different  tensions. 
Denote  these  tensions  by  Sl  (=  P)  £2,  £„  #4,  £.,  #.,  and  #,. 

For  simplicity  (as  will  be  seen)  let  us  suppose  P  given  and  Q 
required.  Since  each  pulley  of  this  figure  is  similarly  circum- 
stanced to  the  one  in  Fig.  15  where  the  tension  on  the  unwinding 

side  was  greater  than  that  on  the  other  in  the  ratio  -  r  —  rr^ 

r  —  (a  -\-  b) 

(where  a,  r,  and  b  have  the  same  meanings  as  in  the  preceding 
paragraph)  we  see  that 


and  so  on,  -up  to 


Hence  we  adopt  the  following  construction  :  layoff  on  a  hori- 
zontal line,  o  .  .  w  =  r  —  (a  -f-  &);  o  .  .wf  =  r  -J-  (a  -j-  l>)  •  and  also 
w'  .  .  of  =  r  —  (a  +5).  Draw  a  vertical  through  each  of  the 
points  o.  w,  </,  and  w'.  On  the  vertical  through  w,  lay  off  to 
scale  w..m  =  P  =  $„  and  draw  and  prolong  m  .  .  o'  to  intersect 
the  vertical  through  w'  in  some  point  m'\  whence  w'  m'  =  /%. 
Then  a  line  joining  m'  with  o,  by  its  intersection  with  the  verti- 
cal through  w,  gives  a  point  n  such  that  w  .  .  n  =  S3  ;  while 
n  .  .  o'  prolonged  cuts  the  vertical  through  w'  in  some  point  TZ/, 
giving  w'ri  =  S4  ;  and  so  on,  until  all  the  six  tensions  St  to  $7 
are  found. 

Proof.  (By  similar  triangles  we  see  that  the  proportions  men- 
tioned above  are  satisfied.) 

We  then  have  (see  [(7]  in  figure), 

Q  =%  +  %+%+%  +  8.+  %;    asalso 


p 

Having  thus  found  the  ratio  of  P  to  Q,  i.e.,  the  value  of  -^r  ' 

V 

we  can  easily  compute  P  if  Q  is  given,  as  in  a  practical  case  ;  and 
also  the  efficiency,  since  with  no  friction  P0  =  ^  Q  (for  the 
tackle  in  this  case). 


21  NOTES  ON  THE   GRAPHICAL   STATICS  OF  MECHANISM. 

In  backward  motion,  having  P  =  Sl  given,  we  would  draw  a 
line  from  m  through  o,  instead  of  through  o',  to  fix  m',  and  re- 
turning, a  line  through  m'  and  o'  to  fix  n,  and  so  on,  each  tension 
so  found  being  greater  than  the  one  preceding  it.  And  finally  we 
would  have  Q  =  S^  -f-  &3  -f-  S4  +  $5  +  Se  -j-  #7>  and  the  ratio  of 
the  given  Q  to  the  unknown  P'  thus  fixed  ;  and  P'  would  be 

7J/  O 

found  from  the  relation  :  -—-=  l 


Q 

21.  Example  X.  The  Differential  Pulley.  (Plate  Y,  Fig.  17.) 
This  is  a  tackle  in  which  the  upper  block  carries  only  one  pulley, 
which,  however,  has  two  grooves  in  ||  planes,  but  with  slightly 
different  radii.  Also,  since  friction  in  the  grooves  is  not  sufficient 
for  the  purpose,  projecting  pegs  or  ridges  (or  some  similar  device) 
are  provided  to  prevent  the  chain  from  slipping  in  either  groove. 
The  lower  block  D  carries  an  ordinary  pulley  of  one  groove,  the 
weight  Q  being  suspended  from  the  axle  of  this  lower  block. 

The  chain  is  endless,  passing  twice  over  the  pulley  A  (once 
in  each  groove)  and  once  under  the  lower  pulley,  while  a  portion 
hangs  freely,  as  shown. 

Given  the  load  Q,  required  the  vertical  force  (or  load  perhaps) 
P,  to  be  applied  to  the  chain  where  it  unwinds  from  the  outside 
groove  (see  figure)  in  order  to  raise  Q  and  overcome  all  friction. 
The  lower  pulley  is  then  acted  on  by  the  vertical  force  Q,  and 
the  two  vertical  and  upward  tensions  S1  and  $,  each  of  these  three 
forces  being  tangent  to  its  own  friction-circle,  as  shown,  on  the 
proper  side  (note  that  $i  is  on  the  unwinding  side).  We  find  Sl 
and  S9  by  treating  Q  as  a  load  (§  329)  resting  on  a  horizontal 
beam  supported  in  verticals  a  and  n.  At  \B\  is  the  force-diagram, 
while  a . .  b . .  n  is  the  equilibrium  polygon.  (See  §  329.) 

/Si  and  S9  having  thus  been  determined,  we  consider  the  upper 
pulley,  which  is  acted  on  by  four  ||  forces,  viz.,  the  known  $,  and 
$2,  the  unknown  .Pand  the  unknown  R  or  reaction  at  the  journal 
of  the  pulley,  The  four  action-lines  are  known,  being  vertical 
and  tangent  to  the  respective  friction-circles  in  the  manner  shown. 
Note  the  direction  of  the  uniform  motion  (to  raise  the  load  Q). 

Again  we  employ  §  329,  regarding  A  as  a  horizontal  beam  or 
lever  with  supports  in  the  verticals,/  and  c,  and  loaded  with  St 


12 


22  NOTES   ON   THE   GRAPHICAL   STATICS   OF   MECHANISM. 


and  SM  both  known.  We  lay  off  r . .  s  =  Sa  and  s . .  t  =  S»  and 
take  any  pole  0,  drawing  rays  from  O  to  r,  s,  and  t.  From  any 
point  G  in  the  action-line  of  R  (the  left-hand  reaction,  or  support- 
ing force)  we  draw  a  line  ||  to  0 . .  rto  find  d,  then^Z .  .e\\toO..s 
to  find  e,  and  a  line  parallel  to  O .  .  t  through  e,  to  find  f  in  the 
action-line  of  the  Tight-hand  supporting  force,  P.  Drawing  c .  .f, 
a  line  parallel  to  it  through  0  fixes  n'  on  the  load-line  (produced), 
giving  in'  —  P,  and  n'w  =  -Z?  ;  i.e.,  \A\  is  the  force-diagram. 

Without  friction,  the  vertical  action-lines  would  be  drawn 
through  the  centres  of  the  friction-circles,  and  anew  construction 
on  this  basis  would  give  P0,  whence  the  efficiency  P0  ~  P  can 
be  found. 

For  backward  motion  each  force-vertical  shifts  over  to  the 
opposite  side  of  the  friction-circle  from  that  shown  in  Fig.  17, 
and  the  result  of  a  third  construction  is  P' .  If  P'  is  found 
to  be  negative,  that  is,  if  n  occurs  above  t  in  diagram  A,  the 
mechanism  is  self-locking^  as  should  be  the  case  in  the  practical 
machine  itself. 

22.  Rigidity  of  Heinp  Eopes.  Here,  as  with  chains,  the  effect  of 
the  rigidity  is  to  cause  the  tension  where  the  rope  is  winding  on 
to  have  a  lever-arm  about  the  centre  of  the  pulley  —  r  -f-  a,  where 
r  =  radius  of  circle  formed  by  the  axis  of  the  rope  when  wound 
on  the  pulley,  and  a  =  a  small  distance  which  from  Eytelwein's 
formula  for  rigidity  of  hemp  ropes  may  be  put  —  0.0093<$2,  where 
d  is  the  diameter  of  the  rope  in  millimetres,  and  whence  a  will 
be  obtained  in  millimetres. 

The  tension  on  the  unwinding  side  has  a  lever-arm  of  r  —  a. 
Hence,  having  computed  a,  we  deal  with  hemp  ropes  as  with 
chains.  The  phenomena  observed  with  wire  ropes  are  different. 
(See  §  176.) 

23.  Tooth    Friction    in    Spur    Gearing.     (Plate  Y,  Fig.  18.) 
This  figure  shows  one  gear-wheel  driving  another,  both  provided 
with  "  involute  teeth "  by  which  we  are  to  understand  that  the 
normal  a^..o,  or  o . .  a»  at  the  point  of  contact  always   passes 
through  o,  the  intersection  of  the  line  of  centres   with  the  pitch- 
circle,  as  motion  proceeds. 

We  assume  here  that  two  pairs  of  teeth  are  always  in  contact. 


23  KOTES    ON    THE    GRAPHICAL    STATICS    OF    MECHANISM. 

Just  now  these  points  of  contact  are  at  av  and  a,,  and  have  there- 
fore a  common  normal  a,0«2. 

Rubbing  occurs  both  at  0;,  and  &2,  and  evidently  in  such  direc- 
tions that  the  pressure  at  a  has  a^  0X  as  action  line  ;  and  that  at  «a 
has  02&2  as  action-line,  making  the  angle  of  friction  with  the 
respective  normals  (or  common  normal,  rather).  This  common, 
normal  a^oa^  will  be  assumed  as  making  an  angle  of  75°  with  the 
line  of  centres  at  all  times  (property  of  the  kind  of  teeth  used). 
Hence  the  resultant  action  of  the  two  driving  teeth  upon  those 
driven  is  represented  by  an  ideal  force  R,  the  resultant  of  the 
pressures  at  al  and  aa,  and  acting  through  0',  making  an  angle  of 
75°  with  the  line  of  centres.  Notice  the  position  of  this  angle 
with  reference  to  the  direction  of  motion  and  to  the  driven  wheel ; 
also  that  the  effect  of  friction  is  to  cause  the  action-line  of  J?, 
which  without  friction  would  act  along  a^oav  to  be  shifted  ||  to 
itself  a  distance  oof  farther  from  the  centre  of  the  driving  wheel. 
This  distance  00',  can  ensily  be  determined  by  drawing  the  parts 
concerned  on  a  convenient  scale,  and  will  be  called  C  in  the  next 
paragraph. 

24.  Example  XI.  Pinion  Spur-Wheel,  Drum  and  Weight. 
(Plate  Y,  Fig.  19.)  The  weight  Q  hangs  by  a  chain  or  rope  from 
the  drum  B  which  forms  a  rigid  body  with  the  spur-wheel  //, 
with  which  the  pinion  A  gears.  A  drives  //,  and  it  is  required  to 
find  what  force  P,  applied  to  a  crank  d  (forming  one  piece  with 
the  pinion)  and  acting  (at  this  instant)  in  the  line  ~b  . .  03,  will  main- 
tain uniform  motion  ;  i.e.,  overcome  all  frictions  and  raise  Q 
without  acceleration. 

Since  A  drives  ZTwith  tooth-gearing  (involute  and  of  same 
design  as  in  preceding  paragraph),  the  line  of  action  of  the  result- 
ant pressure  7?2  between  them  is  00'0:,  making  an  angle  of  75° 
with  the  line  of  centres,  as  shown  (note  on  which  side),  and  is 
drawn  through  the  point  o'  on  the  line  of  centres  but  at  a  distance 
=  C  farther  from  the  centre  of  the  driving  pinion  A  than  a  point 
in  the  pitch-circle  of  the  latter.  Call  this  force  ^?2. 

The  reaction  at  the  bearing  s  is  some  force  Rl  whose  action- 
line  must  pass  through  S,  the  intersection  of  the  action-lines  of 
the  other  two  forces,  7?2  and  P  (since  A  is  a  three-force  piece), 


24  NOTES   ON   THE    GRAPHICAL   STATICS   OF   MECHANISM, 

and  be  tangent  (on  the  right)  to  the  friction-circle  at  s.  The  ac- 
tion-line of  Q  is  vertical,  and  is  tangent  (on  the  left)  to  -a  friction- 
circle  at  a  ( just  as  in  Fig.  15  a  similar  relation  holds  at  K) ;  it 
cuts  g  . .  1}  at  <7,  and  therefore  a  line  drawn  through  g,  and  tangent 
(on  right)  to  the  friction  circle  at  K^  is  the  action-line  of  R^  the 
reaction  at  the  bearing  K. 

We  thus  have  the  action-lines  of  all  three  forces  acting  on' 
each  of  the  three-force  pieces,  A  and  H,  while  the  force  Q  is 
given.  Hence,  the  force-triangle  r..m..n  is  easily  drawn  for 
piece  H,  and  determines  7?2  and  Ry  With  n"m"  =  and  ||  to 
m . .  n  as  a  known  side,  we  then  complete  the  force-triangle  for 
piece  A9  from  which  m" . .  r"  =  P  is  scaled  off. 

Without  friction,  j??2  would  shift  to  the  position  g0 . .  J0,  Rl 
would  pass  through  &0  and  the  centre  of  the  circle  at  s.  R%  would 
pass  through  the  centre  of  the  circle  at  B  and  the  point  gr0,  in  the 
new  vertical  action-line  of  Q  (through  centre  of  friction -circle 
mentioned  above).  g0 . .  t>0  is  parallel  to  g . .  b  and  passes  through 
the  intersection  o  of  the  line  of  centres  c . .  s,  and  the  pitch-circle 
of  the  pinion  A.  Drawing  the  dotted  force-triangles  on  this  basis, 
Q  being  given,  we  finally  obtain  P0  =  m"0. .  r0".  The  efficiency 
can  now  be  obtained,  =P0  -r-  P. 

25.  Belt  Gearing.  In  Fig.  20,  Plate  VI,  we  have  a  pulley 
turning  in  a  fixed  bearing  and  driven  by  a  force  P.  By  belt 
connection  this  pulley  drives  another,  not  shown  in  the  figure. 
The  tension  Sw  on  the  driving  side  is  greater  than  that,  $0,  on  the 
following  side.  If  Z  is  the  (ideal)  resultant  of  Sn  and  80)  then 
the  reaction  R  of  the  bearing  must  act  in  a  line  through  #,  the 
intersection  of  P  and  Z  and  tangent  (above)  to  the  friction-circle 
at  the  bearing  ;  i.e.,  it  acts  along  a.  .  &.*  If  we  assume  that  the 
belt  is  on  the  point  of  slipping  on  the  smaller  of  the  two  pulleys, 
we  have  the  relation 

Sn  =  8.e» (§  170) 

where/"  =  coefficient  of  friction,  e  is  the  Naperian  Base,  and  a  = 
arc  of  contact  on  the  smaller  pulley  in  n  measure,  or  in  radians. 
Although  Sn  and  $0  are  both  unknown  at  the  outset,  we  have 
*  By  mistake,  R  has  been  drawn  in  Fig.  20  along  a' . .  &,  instead  of  a  .  .  b. 


25  NOTES   ON   THE   GRAPHICAL   STATICS   OF   MECHANISM. 

their  ratio  from  the  above  equation,  and  hence  can  construct  the 
action-line  of  Z(iov  impending  slip  only,  it  must  be  remembered), 
their  resultant,  thus  determining  the  point  b  in  Fig.  20  and 
ultimately  R  and  Z,  as  will  be  seen.  With  Z  found,  we  can 
obtain  8n  and  S0.  The  value  of  this  ratio,  efa,  having  been 
computed  for  a  range  of  values  off  and  of  a,  the  results  may  be. 
embodied  graphically  in  the  spirals  shown  in  Fig.  21,  Plate  VI, 
these  being  drawn  in  such  a  way,  all  starting  from  the  point  A 
in  the  circumference  of  the  circle  A .  .  (7,  that  if  OA,  the  radius, 
represent  the  smaller  tension,  SQ,  and  the  special  value  a  —  AOO 
in  any  case  be  laid  off  and  the  radius  OG  produced  till  it  inter- 
sects the  spiral  corresponding  to  the  coefficient  f  proper  to  the 
case  in  hand,  B  being  this  intersection ;  then  OB  =  Sn9  and 
GB  —  Sn  —S0  (which  multiplied  by  the  velocity  of  the  belt 
gives  the  power  transmitted).  Or,  whatever  &0  may  be,  the 
ratio  BO  :  AO  =  the  ratio  Sn  :  £0,  and  may  be  obtained  from 
the  diagram  if  f  and  a  are  given.  [N.B. — Note  carefully  that 

o 

the  relation  -~-  =  efa   only   holds   when    the    belt    is   actually 

slipping  on  the  pulley -rim  (and  then/*  is  the  coefficient  of  fric- 
tion of  motion)  or  is  on  the  point  of  slipping  (and  theny  =  co- 
efficient of  friction  of  rest),  and  is  never  to  be  used  except  for 
those  conditions.  Of  course,  in  most  machinery  impending  slip 
is  to  be  avoided,  and  the  only  use  of  the  above  formula  in  such 
cases  is  to  find  the  ideal  maximum  value,  efa,  for  the  ratio 
&n  '•  £„>  which  the  actual  value  should  not  approach  if  slipping  is 
not  to  occur.  For  the  uniform  motion  of  an  "idle  pulley,"' 
ignoring  axle  friction,  Sn  :  S0  is  always  equal  to  1.00. 

26.  Example  XII.  Brake  Strap  and  Drum.  (Plate  VI,  Fig.  22.) 
A  is  a  lever  with  a  fixed  fulcrum  or  bearing  at  B  and  has 
attached  to  it  both  ends  of  the  belt  or  strap  which  passes  over  a 
pulley  and  serves  as  a  brake  to  prevent  the  acceleration  of  the 
descending  weight  Q.  The  chain  sustaining  Q  unwinds  from 
the  drum  (7,  rigidly  attached  to  the  pulley,  which  turns  on  a 
fixed  bearing  B.  Required  the  proper  force  P,  in  a  given 
action-line  r . .  P,  to  be  applied  to  the  lever  A,  to  preserve  a 
uniform  motion  for  Q  (downward). 


26  NOTES   ON   THE   GKAPHICAL   STATICS   OF   MECHANISM. 

Evidently,  from  the  direction  of  motion,  the  tension  in  the 
strap  at  d  is  the  greater,  =  $„,  and  that  in  the  portion  x . .  e  is 
the  smaller,  =  S0.  Since  in  this  case  there  is  actual  slipping  of 
the  drum  under  the  strap,  the  ratio  Sn  :  <SQ  is  known  from  Fig. 
21,  for  (say)/  =  0.18  and  a  =  f  n  (corresponding  to  270°),  to 
be  2.51.  Hence,  from  the  intersection,  0,  of  the  two  straight 
portions  of  the  belt,  we  lay  off  any  convenient  distances  a..e 
and  a . .  d,  such  that  ad  =  2.51  ae,  and  complete  a  parallelogram 
upon  them  as  sides  ;  then  the  diagonal  a . .  o  is  the  action-line  of 
Z,  the  unknown  resultant  of  Sn  and  S9. 

The  pulley  and  drum,  therefore,  constitute  a  three-force  piece 
acted  on  by  Q  in  the  vertical  line  tangent  (on  the  inside)  to  a 
friction-circle  at  t ;  by  the  (ideal)  Z  in  line  a.  .0 ;  and  by  the 
reaction,  ./?,  of  the  bearing  B,  tangent  (on  left)  to  the  friction- 
circle  and  in  a  line  which  must  pass  through  gf  (intersection  of 
the  other  two  force-lines).  Knowing  Q  and  all  three  action-lines 
mentioned,  the  force-triangle  k . .  m . .  n  determines  R  and  Z;  i.e., 
n .  .Jc  and  m  ..n.  As  for  the  lever  A,  though  there  are  really 
four  forces  acting  on  it,  viz.,  jP,  Sn,  /#„,  and  R^  (the  reaction  at  B\ 
we  may  treat  it  as  a  three-force  piece,  since  Sn  and  S0  are  equiva- 
lent to  the  force  Z,  now  known  both  in  amount  and  position. 
Hence  we  draw  h .  .f  =  and  ||  to  Z,  then  f ..  g\\io  r . .  s  (since 
Zand  P  meet  at  r)  and  h..g  \\  to  r..P,  thus  fixing  j?,  — 
f . .  g  ||  and  P  =  g . .  h.  Then  by  resolving  Z  along  a  . .  d  and 
a . .  e  we  find  Sn  and  /#„.  (Note. — Q  is  the  working  force  in  this 
example,  and  P  neutral.  All  work  is  spent  in  friction.) 

27.  Example  XIII.  Transmission  of  Power  through  two  Pulleys 
having  Belt  Connection.  (Plate  VI.  Fig.  23.)  Let  the  pulley 
A  drive  the  pulley  B  with  uniform  motion.  At  this 
instant  the  useful  resistance  is  Q  acting  on  B  in  line  m . .  q. 
Given  Q,  required  P,  acting  in  line  £ . .  a  on  pulley  J.,  to  over- 
come Q  and  the  journal  frictions  at  B  and  B'.  We  assume  that 
the  belt  is  on  the  point  of  slipping  on  the  circumference  of  the 
smaller  pulley,  B\  then  Sn  =  S0  efa,  where  a  is  the  arc  of  con- 
tact on  the  small  pulley.  Prolong  the  action-lines  of  8n  and  $0 
to  their  intersection  <?.  From  Fig.  22  find  the  ratio  of  Sn  to  St 


27  NOTES   ON   THE   GEAPHICAL   STATICS   OF   MECHANISM. 

for  the  given  coefficient  of  friction  and  the  arc  «,  and  lay  off 
o  . .  I  and  o  . .  Jc  in  this  ratio  (ok  >  ol)  at  convenience.  This  gives 
o .  .i  as  the  action-line  of  Z,  the  (ideal)  resultant  of  Sn  and  $0 ; 
of  course,  that  there  may  be  no  slip  or  any  approach  to  it,  actual 
values  must  be  secured  for  these  tensions  greater  than  those  to 
be  found  by  this  construction,  which  are  for  impending  slip  c/n> 
small  pulley  (and  this  means  the  assumption  of  a  less  value  /or 
the  ratio  Sn  :  &0.  (See  p.  186.) 

A  is  a  three-force  piece  (so  considered  here)  under  the  action 
of  Z  (ideal),  P9  and  J?,  the  bearing  reaction.  Pulley  B  may 
also  be  treated  as  a  three-force  piece  under  action  of  Q,  of  Z 
(reversed)  and  R,  the  bearing  reaction  at  B.  P  and  Z  intersect 
at  Z>,  Q  and  Z  at  q.  Hence  R^  acts  through  ~b  and  tangent  (above) 
to  its  friction-circle ;  while  R  acts  through  q  and  is  tangent  (on 
right)  to  friction -circle  at  B. 

Beginning,  then,  with  pulley  B,  since  the  force  Q  is  given,  we 
close  the  triangle  v . .  w  . .  x  in  an  obvious  manner,  obtaining  R  and 
Z.  For  pulley  A,  now  that  Z  is  found,  we  complete  the  force- 
triangle  y  . .  z  . .  d,  and  determine  R^  and  P. 

Without  friction  at  the  bearings^  R  and  Rl  would  pass  through 
the  centres  of  their  friction-circles  and  the  dotted  force-triangles 

O 

would  result,  whence  we  have  P0  =  z . .  df. 

To  find  the  belt  tensions  (for  impending  slip  on  smaller  pul- 
ley) we  resolve  the  force  Z\\  to  their  directions;  see  lower  part 
of  the  figure. 

Note. — As  far  as  finding  the  value  of  P  alone  is  concerned, 
having  Q  given,  any  line  whatever  could  be  taken  through  the 
point  o,  as  the  action -line  of  the  resultant  of  the  two  tensions,  if 
the  friction  at  the  bearings  were  disregarded,  and  the  construc- 
tion would  result  in  the  same  value  of  jP,  whatever  the  belt- 
tensions,  provided  the  belt  did  not  slip. 

28.  Final  Remark.  From  an  inspection  of  the  preceding  ex^ 
amples  involving  the  effect  of  friction  in  the  working  of  machines, 
it  becomes  apparent,  as  should  be  expected,  of  course,  that  in 
every  case  this  effect  is  to  put  the  working  force  at  the  greatest 
possible  disadvantage,  thus  exacting  as  large  a  value  as  possible 


NOTES   ON  THE   GKAPHICAL   STATICS   OF   MECHANISM.  28 

for  it;  and  from  this  general  principle  we  may  often  decide  quickly 
in  the  matter  of  tangencies  to  friction-circles,  inclination  of  a 
pressure  on  one  side  or  the  other  from  the  normal,  etc. 


Fig.  A. 


pole 


Fig.  D, 


Am.Dk.  Xote  Co.  N.  Y. 


CONTENTS. 


PAGE 

Assumptions 1 

Efficiency 2,  3 

"  Overhauling" 3,  4 

Sliding  Friction 4 

Mill  Elevator 5 

Wedge 6 

Jack-screw 7 

Pivot  and  Journal  Friction 9 

Bell-crank...? 9 

Slider-crank 10 

Beam-engine 11 


Oscillating  Engine 13 

Ore-crusher 14 

Rolling  Friction 15 

Crane  (rollers  of) 17 

Chain  Friction 18 

Tackle 19 

Differential  Pulley 21 

Rigidity  of  Ropes 22 

Spur  Gearing 22 

Belt  Gearing 24 

Brake-strap  and  Drum 25 


PLATE  I.       Figures  1  to  4 . 


Am.  Ek.  Note  Co.  N.  T. 


' '   '  ' 

'<•    J    '       'r   '     1       '  ' 


PLATE  II.       Figures  5  to  8 


Fig,  8. 

THE  "EVANS" 

STRAIGHT-LINE 

MOTION. 


PLATE  m.       Figures  9  to  14. 


Fig.  9. 


OSCILLATING  ENGINE, 


WLU  FRICTION.  _ 


Horizontal  Section  itf  $w\nqing  Qrane  and 
Fixed  Mast,  about  to/itc/V  it  it^nis.'.,  yhis  sectwrj, 
is  at  the  base  Yin  the  Korteontdl pRinz  XY.  •»  J 
The  pivot  frictio^i  al  Tisnqt  considered. 


PLATE  IV.       Figures  15  and  16. 


Am.  Bk.  Note  Co.  N.  Y. 


PLATE  V.       Figures  17  to  19. 


Fig.  17. 

P  DIFFERENTIAL.PULLEY 


PLATE  VI.       Figures.20  to  23, 


Fig,2l,       BELT-FRICTION  SPIRAL 


Fig.  2O.      BELT-FRICTION 


LOGARITHMS  (BRIGGS*). 

N 

01234 

5    G    7    8    9 

Dif. 

1O 

11 
12 
13 

14 

0000  0043  0086  0128  0170 
0414  0453  0492  0531  0569 
0792  0828  0864  0899  0934 
1139  1173  1206  1239  1271 
1461  1492  1523  1553  1584 

0212  0253  0294  0334  0374 
0607  0645  0682  0719  0755 
0969  1004  1038  1072  1106 
1303  1335  1367  1399  1430 
1614  1644  1673  1703  1732 

42 
38 
35 
32 
30 

15 

16 
17 

18 
19 

1761  1790  1818  1847  1875 
2041  2068  2095  2122  2148 
2304  2330  2355  2380  2405 
2553  2577  2601  2625  2648 
2788  2810  2833  2856  2878 

1903  1931  1959  1987  2014 
2175  2201  2227  2253  2279 
2430  2455  2480  2504  2529 
2672  2695  2718  2742  2765 
2900  2923  2945  2967  2989 

23 
26 
25 
24 
22 

2O 

21 
22 
23 
24 

3010  3032  3054  3075  3096 
3222  3243  3263  3284  3304 
3424  3444  3464  3483  3502 
3617  3636  3655  3674  3692 
3802  3820  3838  3856  3874 

3118  3139  3160  3181  3201 
3324  3345  3365  3385  3404 
3522  3541  3560  3579  3598 
3711  3729  3747  3766  3784 
3892  3909  3927  3945  3962 

21 
20 
19 
19 

18 

25 

26 

27 
28 
29 

3979  3997  4014  4031  4048 
4150  4166  4183  4200  4216 
4314  4330  4346  4362  4378 
4472  4487  4502  4518  4533 
4624  4639  4654  4669  4683 

4065  4082  4099  4116  4133 
4232  4249  4265  4281  4298 
4393  4409  4425  4440  4456 
4548  4564  4579  4594  4609 
4698  4713  4728  4742  4757 

17 
16 
16 
15 
15 

3O 

31 
32 
33 
34 

4771  4786  4800  4814  4829 
4914  4928  4942  4955  4969 
5051  5065  5079  5092  5105 
5185  5198  5211  5224  5237 
5315  5328  5340  5353  5366 

4843  4857  4871  4886  4900 
4983  4997  5011  5024  5038 
5119  5132  5145  5159  5172 
5250  5263  5276  5289  5302 
5378  5391  5403  5416  5428 

14 
14 
13 
13 
13 

35 

36 
37 

38 
39 

5441  5453  5465  5478  5490 
5563  5575  5587  5599  5611 
5682-  5694  5705  5717  5729 
5798  5809  5821  5832  5843 
5911  5922  5933  5944  5955 

5502  5514  5527  5539  5551 
5623  5635  5647  5658  5670 
5740  5752  5763  5775  5786 
5855  5866  5877  5888  5899 
5966  5977  5988  5999  6010 

12 
12 
12 
11 
11 

40 

41 
42 
43 

44 

6021  6031  6042  6053  6064 
6128  6138  6149  6160  6170 
6232  6243  6253  6263  6274 
6335  6345  6355  6365  6375 
6435  6444  6454  6464  6474 

6075  6085  6096  6107  6117 
6180  6191  6201  6212  6222 
6284  6294  6304  6314  6325 
6385  6395  6405  6415  6425 
6484  6493  6503  6513  6522 

11 
10 
10 
10 
10 

45 

46 

47 
48 
49 

6532  6542  6551  6561  6571 
6628  6637  6646  6656  6665 
6721  6730  6739  6749  6758 
6812  6821  6830  6839  6848 
6902  6911  6920  6928  6937 

6580  6590  6599  6609  6618 
6675  6684  6693  6702  6712 
6767  6776  6785  6794  6803 
6857  6866  6875  6884  6893 
6946  6955  6964  6972  6981 

10 
9 
9 
9 
9 

5O 

51 
52 
53 
54 

6990  6998  7007  7016  7024 
7076  7084  7093  7101  7110 
7160  7168  7177  7185  7193 
7243  7251  7259  7267  7275 
7324  7332  7340  7348  7356 

7033  7042  7050  7059  7067 
7118  7126  7135  7143  7152 
7202  7210  7218  7226  7235 
7284  7292  7300  7308  7316 
7364  7372  7380  7388  7396 

9 
9 

8 
8 
8 

N.  B.—  Naperian  log  =  Briggs'  log  x  2.302. 
Base  of  Naperian  system  =  e  =  2.71828. 

LOGARITHMS  (BRIGGS'). 

N 

01234 

56789 

Dif. 

55 

56 
57 

58 
59 

7404  7412  7419  7427  7435 
7482  7490  7497  7505  7513 
7559  7566  7574  7582  7589 
7634  7642  7649  7657  7664 
7709  7716  7723  7731  7738 

7443  7451  7459  7466  7474 
7520  7528  7536  7543  7551 
7597  7604  7612  7619  7627 
7672  7679  7686  7694  7701 
7745  7752  7760  7767  7774 

8 
8 
8 
7 
7 

6O 

61 
62 
63 
64 

7782  7789  7796  7803  7810 
7853  7860  7868  7875  7882 
7924  7931  7938  7945  7952 
7993  8000  8007  8014  8021 
8062  8069  8075  8082  8089 

7818  7825  7832  7839  7846 
7889  7896  7903  7910  7917 
7959  7966  7973  7980  7987 
8028  8035  8041  8048  8055 
8096  8102  8109  8116  8122 

7 

7 
7 
7 
7 

65 

66 

67 
68 
69 

8129  8136  8142  8149  8156 
8195  8202  8209  8215  8222 
8261  8267  8274  8280  8287 
8325  8331  8338  8344  8351 
8388  8395  8401  8407  8414 

8162  8169  8176  8182  8189 
8228  8235  8241  8248  8254 
8293  8299  8306  8312  8319 
8357  8363  8370  8376  8382 
8420  8426  8432  8439  8445 

7 
7 
6 
6 
6 

70 

71 

72 
73 

74 

8451  8457  8463  8470  8476 
8513  8519  8525  8531  8537 
8573  8579  8585  8591  8597 
8633  8639  8645  8651  8657 
8692  8698  8704  8710  8716 

8482  8488  8494  8500  8506 
8543  8549  8555  8561  8567 
8603  8609  8615  8621  8627 
8663  8669  8675  8681  8686 
8722  8727  8733  8739  8745 

6 
6 
6 
6 
6 

75 

76 

77 
78 
79 

8751  8756  8762  8768  8774 
8808  8814  8820  8825  8831 
8865  8871  8876  8882  8887 
8921  8927  8932  8938  8843 
8976  8982  8987  8993  8998 

8779  8785  8791  8797  8802 
8837  8842  8848  8854  8859 
8893  8899  8904  8910  8915 
8949  8954  8960  8965  8971 
9004  9009  9015  9020  9025 

6 
6 
6 
6 
5 

80 

81 
82 
83 
84 

9031  9036  9042  9047  9053 
9085  9090  9096  9101  9106 
9138  9143  9149  9154  9159 
9191  9196  9201  9206  9212 
9243  9248  9253  9258  9263 

9058  9063  9069  9074  9079 
9112  9117  9122  9128  9133 
9165  9170  9175  9180  9186 
9217  9222  9227  9232  9238 
9269  9274  9279  9284  9289 

5 
5 
5 
5 
5 

85 

86 
87 
88 
89 

9294  9299  9304  9309  9315 
9345  9350  9355  9360  9365 
9395  9400  9405  9410  9415 
9445  9450  9455  9460  9465 
9494  9499  9504  9509  9513 

9320  9325  9330  9335  9340 
9370  9375  9380  9385  9390 
9420  9425  9430  9435  9440 
9469  9474  9479  9484  9489 
9518  9523  9528  9533  9538 

5 
5 
5 
5 
5 

9O 

91 
92 
93 

94 

9542  9547  9552  9557  9562 
9590  9595  9600  9605  9609 
9638  9643  9647  9652  9657 
9685  9689  9694  9699  9703 
9731  9736  9741  9745  9750 

9566  9571  9576  9581  9586 
9614  9619  9624  9628  9633 
9661  9666  9671  9675  9680 
9708  9713  9717  9722  9727 
9754  9759  9763  9768  9773 

5 
5 
5 
5 
5 

95 

96 

97 
98 
99 

9777  9782  9786  9791  9795 
9823  9827  9832  9836  9841 
9868  9872  9877  9881  9886 
9912  9917  9921  9926  9930 
9956  9961  9965  9969  9974 

9800  9805  9809  9814  9818 
9845  9850  9854  9859  9863 
9890  9894  9899  9903  9908 
9934  9939  9943  9948  9952 
9978  9983  9987  9991  9996 

5 

4 
4 

4 

4 

N.  B.  —  Naperian  log  =  Briggs'  log  x  2.302. 
Base  of  Naperian  System  =  e  —  2.71828. 

TRIGONOMETRIC  RATIOS  (Natural);  including  "  arc,"  by  which  is  meant  the  •  '  it- 

measure"  or  "circular  measure"  of  the  angle;  e.g.,  arc  100°  =  1.7453293,  =  j-f$of  ie. 

arc 

degr. 

sin              esc 

tan              cot 

sec                 cos 

O.OOO 

O 

o.ooo         inf. 

o.ooo         inf. 

I.OOO          I.OOO 

90 

L57I 

O.OI7 

I 

0-017     57-3 

0.017     57-3 

I.OOO           I.OOO 

89 

T.553 

0.035 

2 

0.035     28.7 

0.035     28.6 

i.  ooi        0.999 

88 

1.536 

0.052 

3 

0.052     19.1 

0.052     19.1 

i.ooi        0.999 

87 

1.518 

0.070 

4 

0.070     14.3 

0.070     14.3 

I.OO2           0.998 

86 

1.501 

0.087 

5 

0.087     IJ«5 

0.087      IX«4 

1.004       0.996 

85 

1.484 

0.105 

6 

0.105       9«6 

0.105        9.5 

1.  006       0.995 

84 

1.466 

0.122 

7 

O.I  22           8.2 

0.123       8.1 

1.008       0.993 

83 

1.449 

0.139 

8 

0.139           7-2 

0.141        7.1 

i.oio       0.990 

82 

1.432 

0.157 

9 

0.156          6.4 

0.158       6.3 

1.  012           0.988 

81 

1.414 

0.174 

10 

0.174          5.8 

0.176       5.7 

I.OI5           0.985 

80 

1.396 

0.192 

ii 

O.I9I           5.24 

0.194       5.14 

I.OI9           0.982 

79 

1.379 

0.209 

12 

0.208       4.81 

0.213       4.70 

I.O22           0.978 

78 

1.361 

O.227 

13 

0.225       4-45 

0.231        4-33 

1.026           0.974 

77 

1.344 

0.244 

14 

0.242       4.13 

0.249        4«01 

I.03I           0.970 

76 

1.326 

O.262 

*5 

0.259       3-86 

0.268        3.73 

1.035           0.966 

75 

1.309 

0.279 

1  6 

0.276       3.63 

0.287        3.49 

1.040           0.961 

74 

1.291 

0.297 

17 

0.292       3.42 

0.306        3.27 

1.046           0.956 

73 

1.274 

0.314 

18 

0.309       3.24 

0.325        3.08 

I.05I           0.951 

72 

1.257 

0.332 

I9 

0.326       3.07 

0.344        2.90 

1.058           0.946 

7i 

1.239 

0-349 

20 

0.342       2.92 

0.364        2.75 

1.064           0.940 

70 

1.222 

0.366 

21 

0.358       2.790 

0.384        2.605 

I.07I           0.934 

69 

I.2O4 

0.384 

22 

0.375        2.669 

0.404        2.475 

1.079           0.927 

68 

I.I87 

0.4OI 

23 

0-39i        2.559 

0.424        2.356 

1.  086           0.921 

67 

1.169 

0.419 

24 

0.407        2.459 

0.445        2.246 

1.095        °«9I4 

66 

1.152 

0.436 

25 

0.423       2.366 

0.466        2.145 

1.103        0.906 

65 

LI34 

0-454 

26 

0.438        2.281 

0.488        2.050 

1.113        0.899 

64 

I.II7 

0.471 

27 

0.454       2.203 

0.510        1.963 

1.  122           0.891 

63 

1.099 

0.489 

28 

0.469        2.130 

0.532        i.88r 

I.I33           0.883 

62 

1.082 

0.506 

29 

0.485        2.063 

0.554        1.804 

T.I43           0.875 

61 

1.064 

0.523 

30 

0.500           2.000 

0.577        1-732 

1.155        0.866 

60 

1.047 

0.541 

31 

0.515              .942 

0.601        1.664 

1.167        0.857 

59 

1.030 

0.558 

32 

0.530              .887 

0.625        i.  600 

1.179       0.848 

58 

1.  012 

0.576 

33 

0-545              -836 

0.649        I-54° 

1.192        0.839 

57 

0-995 

0-593 

34 

0.559             .788 

0.675        1.483 

1.206       0.829 

56 

0.977 

0.611 

35 

0-574         -743 

0.700       1.428 

1.  221           0.8l9 

55 

0.960 

0.628 

36 

0.588         .701 

0.727        1.376 

1.236           0.809 

54 

0.942 

0.646 

37 

0.602         .662 

0-754       1-327 

1.252           0.799 

53 

0.925 

0.663 

38 

0.616         .624 

0.781        1.280 

1.269           0.788 

S2 

0.908 

0.68  1 

39 

0.629         -589 

0.810        1.235 

1.287           0.777 

51 

0.890 

0.698  , 

40 

°-643         -556 

0.839        1-192 

1.305           0.766 

5° 

0.873 

0.716      41 

0.656         .524 

0.869       1-150 

1.325           0*755 

49 

0.855 

0-733 

42 

0.669         -494 

0.900       i.  in 

1.346           0.743 

48 

0.838 

0.750 

43 

0.682         .466 

0.933       1-072 

1.367           0.731 

47 

0.820 

0.768 

44 

0.695          -44° 

0.966       1.036 

1.390           0.719 

46 

0.803 

0-785 

45 

0.707          .414 

I.OOO           1.  000 

I.4I4          0.707 

45 

0.785 

cos                sec 

cot               tan 

csc                sin 

degr. 

arc 

'67  -S  PM 


FEJ 


ocr 


AUQ   18  1947 


LD 


U.  C.  BERKELEY  LIBRARIES 


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